题目：
统计各位数字之和是5的数
本题要求实现两个函数：一个函数判断给定正整数的各位数字之和是否等于5；另一个函数统计给定区间内有多少个满足上述要求的整数，并计算这些整数的和。 我讲得可能不算很清楚，具体请查看题目链接 代码实现：
#include<stdio.h> #include<math.h> #define N 5 #define M 10 int cal(int n){ int m = 0; for (;;n/=10) { if (n)m++; else break; } return m; } int judge(int i,int con) { int num = cal(i), sum = 0, q = i; for (int j = num; j > 0; j--) { //calculate the sum of decimal digits of this number int tem= q / pow(M, j - 1); sum += tem; q -= tem*pow(M, j - 1); } if (sum == con)return 1; else return 0; } void sta(int a, int b, int n,int count) { int add = 0; for (int i = a; i <= b; i++) { if (judge(i,N)) { count++; add += i; } } printf("count is %d,sum is %d\n", count,add); } int main() { int a, b, n, count = 0; //Calculate the sum of all numbers which could meet the condition scanf("%d,%d", &a, &b); sta(a, b, N, count); //Judge whether the number you inputed could meet the condition scanf("%d", &n); if (n >= a && n <= b){ if (judge(n, N))printf("%d is counted.\n", n);} else printf("error input!\n"); return 0; }核心函数是judge子函数：
int judge(int i,int con) { int num = cal(i), sum = 0, q = i; for (int j = num; j > 0; j--) { //calculate the sum of decimal digits of this number int tem= q / pow(M, j - 1); sum += tem; q -= tem*pow(M, j - 1); } if (sum == con)return 1; else return 0; }运行效果是这样的（举例）： 或者： 或者：欢迎讨论。
