Acwing 800. 数组元素的目标和

题目描述

代码展示

#include

using namespace std;

const int N = 1e5 + 10;

int a[N], b[N];

int main()
{
    int n , m , x, i = 0, j = m - 1;
    cin >> n >> m >> x;
    
    for (int i = 0; i < n; i ++ ) scanf("%d", &a[i]);
    for (int j = 0; j < m; j ++ ) scanf("%d", &b[j]);
    
    for( i = 0, j = m - 1; i < n; i++){
        while(j >= 0 && a[i] + b[j] > x) j --;
        if(j >= 0 && a[i] + b[j] == x) printf("%d %d\n", i, j);
    }
    return 0;
}