...能做多少就多少吧 2.输入2个整数m、n(m <= n),编答:第二题 include<stdio.h> int Sum(int m,int n);int main(){ int m,n,count;printf("输入两个数\n");scanf("%d%d",&m,&n);count=Sum(m,n);printf("在%d和%d之间所有偶数的和为%d\n",m,n,count);return 0;} int Sum(int m,int n){ int count=0,i;for(i=m;i<=n;i...