回答編集履歴
2
間違いの修正
test
CHANGED
@@ -2,14 +2,14 @@
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---
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- 下一桁9,3,1のみ大きいものから配列に格納する
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- ループ範囲の桁を半分にする
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変更前:999999>>>900000
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変更後:999999>>>999000
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- 1つ目の発見が最大数と仮定し、そこで処理中断することで大幅に改善
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---
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@@ -18,9 +18,53 @@
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---
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最終結果 91 * 99 = 9009
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2桁 1ms
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最終結果 913 * 993 = 906609
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3桁 0ms
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最終結果 9901 * 9999 = 99000099
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4桁 1ms
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最終結果 99681 * 99979 = 9966006699
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5桁 1ms
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最終結果 999001 * 999999 = 999000000999
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6桁 6ms
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最終結果 9997647 * 9998017 = 99956644665999
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7桁 66ms
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最終結果 99990001 * 99999999 = 9999000000009999
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8桁
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8桁 249ms
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-
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最終結果 999920317 * 999980347 = 999900665566009999
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9桁
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9桁 20197ms
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@@ -40,7 +84,7 @@
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```java
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for (int i =
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for (int i = 2; i <= 9; i++) {
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@@ -50,30 +94,6 @@
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func((long) i);
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// func((long)2);//0ms
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// func((long)3);//1ms
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// func((long)4);//1ms
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// func((long)5);//23ms
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// func((long)6);//18ms
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// func((long)7);//1,058ms
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// func((long)8);//994ms
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// func((long)9);//101,748ms
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//最終経過 999920317 * 999980347 = 999900665566009999
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//9桁 101748ms
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long end = System.currentTimeMillis();
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@@ -90,6 +110,8 @@
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```java
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static void func(long num) {
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@@ -104,9 +126,9 @@
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//ループ範囲最大数:3桁例:999999
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final long
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final long iMax = (long) Math.pow(10, num) - 1;
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//ループ範囲最小数:3桁例:999000
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@@ -114,129 +136,67 @@
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//※暫定的に半分にしてみる
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//
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long
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//
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//4 9999 9993 9991 9989 9983 9981
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//30 9909 9903 9901
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//300 9009 9003 9001 9009 9003 9001
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//※全件対象の最大まで配列を確保すると配列数が10倍づつ増えるので
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//※暫定的に100個としておく
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//回文数生成
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final long pd = getPalindrome(i);
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//3桁の整数B:ループ範囲:i>>>最小値√回分数
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long jMax = iMax[n];
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long jMin = (long) Math.sqrt(pd);
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for (long j = jMax; j >= jMin; j--) {
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//整数で割り切れ且つ3桁の整数
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if (pd % j == 0 && pd / j < j) {
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//暫定最大数として更新
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if (maxS < pd) {
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maxA = (int) (pd / j);
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maxB = (int) j;
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maxS = pd;
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break;
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}
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final long iMin = iMax - (long) Math.pow(10, num - (num * 50 / 100)) + 1;
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for (long i = iMax; i >= iMin; i--) {
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//回文数生成
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final long pd = getPalindrome(i);
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//3桁の整数B:ループ範囲:i>>>最小値√回分数
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long jMax = iMax;
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long jMin = (long) Math.sqrt(pd);
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for (long j = jMax; j >= jMin; j--) {
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//整数で割り切れ且つ3桁の整数
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if (pd % j == 0 && pd / j < j) {
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//最初の発見が最大数と仮定する
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if (maxS < pd) {
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maxA = (int) (pd / j);
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maxB = (int) j;
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maxS = pd;
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System.out.println("最終結果 " + maxA + " * " + maxB + " = " + maxS);
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//発見と同時に処理中断で大幅に改善
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return;
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}
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}
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} //j
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} //i
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} //n
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System.out.println("最終経過 " + maxA + " * " + maxB + " = " + maxS);
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}
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@@ -256,8 +216,6 @@
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}
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}
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```
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1
間違いがあったので修正
test
CHANGED
@@ -122,21 +122,25 @@
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//※開始から発見が近いことがほとんどなので
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//※暫定的に1
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//※暫定的に51個としておく
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//※(桁が増えると正確では無いかも)
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long[] iMax = new long[1
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long[] iMax = new long[51];
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int tmp = 0;
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for (int i = 0; i <
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for (int i = 0; i < iMax.length; i=i+3) {
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iMax[i] = startMax -
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iMax[i] = startMax - tmp * 10;
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iMax[i + 1] = startMax -
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iMax[i + 1] = startMax - tmp * 10 - 6;
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iMax[i + 2] = startMax -
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iMax[i + 2] = startMax - tmp * 10 - 8;
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tmp++;
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}
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