编写一个求二维数组主对角线元素和的函数sum( ),在主函数中初始化数组...答:include <iostream>using namespace std;int sum(int a[][4]){int i,j,s=0;for(i=0;i<4;i++)for(j=0;j<4;j++)if(i==j) s+=a[i][j];return s;}int main(){int a[][4]={11,22,33,44,25,35,45,55,18,28,38,48,55,66,77,88};cout<<sum(a);return 0;} ...
求算法,将N个整数分到M个数组中,要求元素和相差最小,元素个数相差最小...答:有N个大小不一的整数,可以是0,将它们分配到M个数组中,要求M数组的元素的和相差最小,数组的元素个数相差最小。比如将{1,2,3,4,5,6,7,8,9,10}分配到4个数组中,其结果可以是 n1 n2 n3 n4 --- 2 1 4 3 5 6 ... 展开 首先要满足的是数组间...
定义一个函数求一维数组中两个元素的和,具体怎么写啊,急答:include<stdio.h>#define N 10int msum(int *a,int n1,int n2) { int x; x=a[n1]+a[n2]; return x; }void main() { int a[N],i,n1,n2,s; for ( i=0;i<N;i++ ) scanf("%d",&a[i]); scanf("%d%d",&n1,&n2); s=msum(a,n1,n2); printf("a[%d]+a[...