1428 - C语言-对角求和

1 2 3
1 1 1
3 2 1

3 7

Solution C

int main(int argc, char* argv[])
{
	int num[3][3],i,j,sumi,sumj;
	sumi=sumj=0;
	for(i=0;i<3;i++)
		for(j=0;j<3;j++)
		{
			scanf("%d",&num[i][j]);
			if(i==j) sumi+=num[i][j];
			if(i==2-j) sumj+=num[i][j];
		}
  printf("%d %d\n",sumi,sumj);
	return 0;
}

Solution C++

#include<iostream>
using namespace std;
int a[4][4];
int main()
{
	for (int i=1; i<=3; i++)
		for (int j=1; j<=3; j++) cin>>a[i][j];
	int sum1=0,sum2=0;
	for (int i=1; i<=3; i++)
	{
		sum1+=a[i][i];
		sum2+=a[i][4-i];
	}
	cout<<sum1<<" "<<sum2<<endl;
	return 0;
}