回答編集履歴
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追記
test
CHANGED
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@@ -5,3 +5,87 @@
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2.1 count[ch-'A']を+1する
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3. これで度数分布表ができた。count[n]の大きいものを出力する。
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[追記] やってみた
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```C
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#include <stdio.h>
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// 度数分布表をcount[]に求める
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void hist(const char* str, int count[26]) {
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int i;
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for ( i = 0; i < 26; ++i ) {
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26
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count[i] = 0;
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}
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while ( *str != '\0' ) {
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int ch = *str - 'A';
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if ( ch >= 0 && ch < 26 ) {
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++count[ch];
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}
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++str;
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}
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}
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int main(){
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int count[26];
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int i;
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int max_inx = 0;
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hist("PPAPAABCD", count);
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// 出現頻度最大のものを探す
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for( i = 0; i < 26; ++i ) {
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if ( count[i] > max_inx ) {
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max_inx = count[i];
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}
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}
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// 出現頻度最大のものを出力する
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for( i = 0; i < 26; ++i ) {
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if ( count[i] == max_inx ) {
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printf("%c : %d\n", i+'A', count[i]);
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}
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}
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return 0;
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}
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```
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