pat 乙级 1021. 个位数统计 (15)

思路 ：

很简单的一道题，建立一个人长度为10 的数组 分别存放0-9的个数  最后遍历 这个数组 不等于0 的输出即可；

代码实现：

#include <stdio.h>#include <iostream>#include <cstdlib>#include <cmath>#include <cctype>#include <string>#include <cstring>#include <algorithm>#include <ctime>using namespace std;char num[1000];int c[10]={0};int main() {gets(num);int len=strlen(num);for (int i = 0; i < len; ++i){if (num[i]=='0'){c[0]++;}if (num[i]=='1'){c[1]++;}if (num[i]=='2'){c[2]++;}if (num[i]=='3'){c[3]++;}if (num[i]=='4'){c[4]++;}if (num[i]=='5'){c[5]++;}if (num[i]=='6'){c[6]++;}if (num[i]=='7'){c[7]++;}if (num[i]=='8'){c[8]++;}if (num[i]=='9'){c[9]++;}}for (int i=0;i<10;i++) {if (c[i]!=0) {cout<<i<<":"<<c[i]<<endl;}}return 0;}

AC图