https://pintia.cn/problem-sets/994805342720868352/problems/994805432256675840
Eva loves to collect coins from all over the universe, including some other planets like Mars. One day she visited a universal shopping mall which could accept all kinds of coins as payments. However, there was a special requirement of the payment: for each bill, she could only use exactly two coins to pay the exact amount. Since she has as many as 10?5?? coins with her, she definitely needs your help. You are supposed to tell her, for any given amount of money, whether or not she can find two coins to pay for it.
Each input file contains one test case. For each case, the first line contains 2 positive numbers: N (≤10?5??, the total number of coins) and M (≤10?3??, the amount of money Eva has to pay). The second line contains N face values of the coins, which are all positive numbers no more than 500. All the numbers in a line are separated by a space.
For each test case, print in one line the two face values V?1?? and V?2?? (separated by a space) such that V?1??+V?2??=M and V?1??≤V?2??. If such a solution is not unique, output the one with the smallest V?1??. If there is no solution, output No Solution instead.
8 15
1 2 8 7 2 4 11 15
4 11
7 14
1 8 7 2 4 11 15
No Solution
代码：
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
int n, target;
int num[maxn];
int main() {
scanf("%d%d", &n, &target);
for(int i = 0; i < n; i ++)
scanf("%d", &num[i]);
sort(num, num + n);
int l = 0, r = n - 1;
int ans1 = -1, ans2 = -1;
while(l < r) {
if(num[l] + num[r] == target) {
ans1 = l;
ans2 = r;
break;
}
if(num[l] + num[r] < target) l ++;
if(num[l] + num[r] > target) r --;
}
if(ans1 == -1 && ans2 == -1)
printf("No Solution\n");
else
printf("%d %d\n", num[ans1], num[ans2]);
return 0;
}
双指针 突然觉得刷 Leetcode 还是很有用的
原文：https://www.cnblogs.com/zlrrrr/p/10119148.html