《算法笔记》 3.1小节-入门模拟->简单模拟问题E 超时解决

很不理解这道题为什么超时。

原代码：

#include<iostream>
#include<math.h>
#include<string>
#include <iomanip> 
#include<stdio.h>
using namespace std;
 
int main()
{
    int n;
    cin >> n;
    int *d = new int[n+1];
    d[0] = 0;
    int s = 0;
    for (int i = 1; i <=n; i++) {
        cin >> d[i];
        s += d[i];
        d[i] = s;
    }
 
    int m;
    cin >> m;
    int x, y;
    for (int i = 1; i <=m; i++) {
        cin >> x >> y;
        int sum = 0;
        int sum2 = 0;
        if (x > y) {
            int tmp = x;
            x = y;
            y = tmp;
        }
        sum = d[y-1] - d[x-1];
        sum2 = s - sum;
        if (sum > sum2)sum = sum2;
        cout << sum << endl;
    }
 
    return 0;
}

就是超时，然后把for循环换成while,ac了。

#include<iostream>
#include<math.h>
#include<string>
#include <iomanip> 
#include<stdio.h>
using namespace std;

int main()
{
	long n;
	cin >> n;
	long *d = new long[n+1];
	d[0] = 0;
	long s = 0;
	for (long i = 1; i <=n; i++) {
		cin >> d[i];
		s += d[i];
		d[i] = s;
	}

	long m;
	cin >> m;
	long x, y;
	while(m--) {//这里
		cin >> x >> y;
		long sum = 0;
		long sum2 = 0;
		if (x > y) {
			long tmp = x;
			x = y;
			y = tmp;
		}
		sum = d[y-1] - d[x-1];
		sum2 = s - sum;
		if (sum > sum2)sum = sum2;
		cout << sum << endl;
	}
	system("pause");
	return 0;
}

还有就是，一开始没想到可以直接叠加距离，这样就不用再用一个For循环算x到y的距离了。