输入NxM矩阵,矩阵元素均为整数,计算其中大于零的元素之和。
#include<stdio.h>
int main()
{
int arr[100] = {0},n,m,sum = 0;
scanf("%d %d",&n,&m);
for(int i = 0;i < n*m;i++)
{
scanf("%d",&arr[i]);
if(arr[i] > 0)
sum += arr[i];
}
printf("%d\n",sum);
return 0;
} #include<iostream>
using namespace std;
int main() {
int N,M,sum;
cin >> N;
cin >> M;
int a[N][M];
for(int i = 0;i<N;i++)
for(int j = 0;j<M;j++){
cin>>a[i][j];
}
for(int i = 0;i<N;i++)
for(int j = 0;j<M;j++){
if(a[i][j]>0){
sum += a[i][j];
}
}
cout<<sum;
return 0;
} import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int num=0;
int m=sc.nextInt();
int n=sc.nextInt();
int[][] array=new int[m][n];
for (int i=0;i<m;i++){
for (int j=0;j<n;j++){
array[i][j]=sc.nextInt();
if(array[i][j]>0){
num+=array[i][j];
}
}
}
System.out.println(num);
}
} #include <stdio.h>
#include <stdlib.h>
int main(void)
{
int sum = 0;
int i, j, n, m;
while (scanf("%d %d", &n, &m) != EOF && (n > 0 && n <= 10) && (m > 0 && m <= 10))
{
int (*a)[m] = (int (*)[m])malloc(n * m * sizeof(int));
if (NULL == a)
{
fprintf(stderr, "Memory allocation failed!\n");
exit(EXIT_FAILURE);
}
for (i = 0; i < n; i++)
{
for (j = 0; j < m; j++)
{
scanf("%d", &a[i][j]);
}
}
for (i = 0; i < n; i++)
{
for (j = 0; j < m; j++)
{
if (a[i][j] > 0)
{
sum += a[i][j];
}
}
}
printf("%d\n", sum);
free(a);
a = NULL;
}
return 0;
} //simple problem;#include <stdio.h>
int main() {
int m,n,a[110],b=0,sum=0;
scanf("%d %d",&m,&n);
while (scanf("%d", &a[b]) != EOF){
if(a[b]>0)sum+=a[b];
b++;
}
printf("%d",sum);
return 0;
} #include <stdio.h>
int main(){
int arr[10][10];
int n,m,i,j;
int sum=0;
scanf("%d%d",&n,&m);
for(i=0;i<n;i++){
for(j=0;j<m;j++){
scanf("%d",&arr[i][j]);
if(arr[i][j]>0){
sum+=arr[i][j];
}
}
}
printf("%d",sum);
return 0;
} #include <iostream>
using namespace std;
int main() {
int n,m,a,b=0;
cin>>n>>m;
for(int i = 0; i< n*m; i++){
cin>>a;
if(a>0){
b+=a;
}
}
cout<<b;
return 0;
}
// 64 位输出请用 printf("%lld")