回答編集履歴
1
コード、実行結果追加
test
CHANGED
@@ -3,3 +3,457 @@
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よってマルチスレッドプログラミングでは、スケジューリングに依存しない書き方が必要で、どうしても実行順序を決めたいときはセマフォなど「同期オブジェクト」を用いてスレッド同士の「待ち合わせ」を行うことが通例です。
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追加
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セマフォ2つを使った同期を示します。
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そこそこまともに動きます。
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```C
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#include <stdio.h>
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#include <time.h>
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#include <stdlib.h>
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#include <string.h>
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#include <pthread.h>
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#include <semaphore.h>
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struct ThreadArgs
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{
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int *count_num;
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int *n;
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int *sum;
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int thread_id;
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sem_t* sem4n;
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sem_t* sem4sum;
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};
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void *thread_func(void *threadArgs);
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void *thread_func(void *threadArgs) {
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int i, thread_id;
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int *n, *count_num, *sum;
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sem_t* sem4n;
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sem_t* sem4sum;
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pthread_detach(pthread_self());
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struct ThreadArgs *args = (struct ThreadArgs *) threadArgs;
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n = args -> n;
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sum = args -> sum;
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count_num = args -> count_num;
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thread_id = args -> thread_id;
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sem4n = args -> sem4n;
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sem4sum = args -> sem4sum;
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free(threadArgs);
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while ( *sum < *count_num) {
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(*n)++;
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printf("Thread%d %d\n", thread_id, *n);
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/* 共有データ*nを使用したので、セマフォの加算 */
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sem_post(sem4n);
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for (i = 0; i < 50000000; i++) {}
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/* 共有データ*sum, *count_numが使えるようになるまで、セマフォの値によるブロック */
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sem_wait(sem4sum);
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}
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}
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int main(int argc, char *argv[])
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{
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int i, j, *n, *top, count_num, thread_num, sum;
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clock_t time1, time0;
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double diff;
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struct ThreadArgs *threadArgs;
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pthread_t mythread;
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int thread_id;
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static sem_t sem4n;
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static sem_t sem4sum;
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if (argc != 3) {
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exit(1);
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}
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count_num = atoi(argv[1]);
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thread_num = atoi(argv[2]);
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printf("begin %d %d\n", count_num, thread_num);
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n = (int *)malloc(sizeof(int) * thread_num);
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/* セマフォの初期化 */
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if (sem_init( &sem4n, 0, 0) != 0) {
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fprintf(stderr, "sem_init() failed\n");
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exit(1);
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}
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if (sem_init( &sem4sum, 0, 0) != 0) {
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fprintf(stderr, "sem_init() failed\n");
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exit(1);
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}
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top = n;
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for (i = 0, n = top; i < thread_num; i++, n++) {
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*n = 0;
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}
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sum = 0;
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time0 = clock();
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thread_id = 0;
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for (i = 0, n = top; i < thread_num; i++, n++) {
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if((threadArgs = (struct ThreadArgs *)malloc(sizeof(struct ThreadArgs))) == NULL){
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fprintf(stderr, "malloc failed?n"), exit(1);
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}
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threadArgs -> count_num = &count_num;
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threadArgs -> n = n;
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threadArgs -> sum = ∑
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threadArgs -> thread_id = thread_id;
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threadArgs -> sem4n = &sem4n;
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threadArgs -> sem4sum = &sem4sum;
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if(pthread_create(&mythread, NULL, thread_func, (void *) threadArgs) != 0){
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fprintf(stderr, "pthread_create() failed\n"), exit(1);
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}
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thread_id++;
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}
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while (sum < count_num) {
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for (i = 0, sum = 0, n = top; i < thread_num; i++, n++) {
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/* 共有データ*nが使えるようになるまで、セマフォの値によるブロック */
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sem_wait(&sem4n);
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sum += *n;
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/* 共有データsum, count_numを使用したので、セマフォの加算 */
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sem_post(&sem4sum);
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}
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//printf("%d %d\n", sum, count_num);
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}
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/* セマフォの削除 */
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sem_destroy(&sem4n);
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sem_destroy(&sem4sum);
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printf("sum %d\n", sum);
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time1 = clock();
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diff =(double)(time1 - time0) / CLOCKS_PER_SEC;
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printf("%.3f[sec]\n", diff);
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return 0;
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}
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```
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> begin 1000 20
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Thread0 1
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Thread1 1
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Thread2 1
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Thread3 1
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Thread4 1
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Thread5 1
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Thread6 1
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Thread7 1
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Thread8 1
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Thread9 1
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Thread10 1
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Thread11 1
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Thread12 1
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Thread13 1
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Thread14 1
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Thread15 1
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Thread16 1
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Thread17 1
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Thread18 1
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Thread19 1
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Thread0 2
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Thread2 2
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Thread5 2
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350
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Thread1 2
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Thread3 2
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354
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Thread6 2
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356
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Thread4 2
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Thread7 2
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360
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Thread9 2
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362
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Thread8 2
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364
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Thread10 2
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366
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Thread11 2
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368
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Thread12 2
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370
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Thread14 2
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Thread13 2
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374
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Thread15 2
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Thread16 2
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Thread17 2
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Thread18 2
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Thread2 3
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384
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385
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Thread6 3
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386
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387
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Thread10 3
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388
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389
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Thread5 3
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390
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391
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Thread0 3
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392
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393
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===中略===
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395
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Thread5 61
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396
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397
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Thread1 61
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398
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+
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399
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Thread9 61
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400
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+
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401
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+
Thread17 40
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402
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+
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403
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Thread13 61
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404
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+
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405
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Thread0 52
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406
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+
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407
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Thread4 52
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408
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+
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409
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+
Thread8 52
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410
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+
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411
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+
Thread12 52
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412
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+
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413
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+
Thread2 54
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414
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+
|
415
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+
Thread16 52
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416
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+
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417
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+
Thread18 51
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418
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+
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419
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Thread6 54
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420
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+
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421
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+
Thread5 62
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422
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+
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423
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+
Thread10 54
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424
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+
|
425
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+
Thread1 62
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426
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+
|
427
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+
Thread14 54
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428
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+
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429
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+
Thread9 62
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430
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+
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431
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+
Thread13 62
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432
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+
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433
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+
Thread19 40
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434
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+
|
435
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+
Thread0 53
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436
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+
|
437
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+
Thread3 41
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438
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+
|
439
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+
Thread4 53
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440
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+
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441
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+
Thread7 41
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442
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+
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443
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+
Thread8 53
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444
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+
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445
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+
sum 1010
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446
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+
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447
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+
128.482[sec]
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448
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+
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449
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+
|
450
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+
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451
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+
環境
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452
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+
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453
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+
Windows 7 Professional SP1
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454
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+
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455
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+
Intel Core i5 750
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456
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+
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457
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+
Cygwin64
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458
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+
|
459
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+
GCC 5.4.0
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