[PAT A1048]Find Coins_综合

[PAT A1048]Find Coins

题目描述

1048 Find Coins （25 分)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.

输入格式

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.

输出格式

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.

输入样例1

8 15
1 2 8 7 2 4 11 15

输出样例1

4 11

输入样例2

7 14
1 8 7 2 4 11 15

输出样例2

No Solution

解析

这道题的大意是，首先输入n和m，n表示硬币的总数，m表示总和，然后输入n个正整数，表示有n个硬币，我们需要做的就是判断有没有两枚硬币加起来之和为m，如果有多组，用v1,v2表示这两个数，要求v1<v2的时候，并且给出最小的v1的那组数据。
这道题目是散列题，不过可以通过多种方式实现，我把所有的方法都给出来希望读者看完后能够自己动手把它敲出来。
散列：

#include<iostream>
#include<string>
#include<map>
using namespace std;
int main()
{
    int n, all;map<int, int> mapp;scanf("%d %d", &n, &all);for (int i = 0; i < n; i++) {
    int temp;scanf("%d", &temp);mapp[temp]++;}bool judge = false;for (auto it : mapp) {
    int a = it.first;int b = all - it.first;if (mapp.find(b) != mapp.end()) {
    if (a == b &&it.second == 1) continue;else {
    judge = true;printf("%d %d", a, b);break;}}}if (!judge) printf("No Solution");return 0;
}

通过自己写二分搜索查找：

#include<iostream>
#include<string>
#include<algorithm>
using namespace std;
int num[100010] = {
     0 };
int main()
{
    int n, all;scanf("%d %d", &n, &all);for (int i = 0; i < n; i++) {
    int temp;scanf("%d", &num[i]);}sort(num, num + n);bool judge = false;for (int i = 0; num[i] <= all - num[i]; i++) {
    int des = all - num[i];int left = i + 1, right = n;while (left <= right) {
    int mid = (left + right) / 2;if (num[mid] == des) {
    judge = true;printf("%d %d", num[i], des);break;}else if (des < num[mid]) {
    right = mid - 1;}else left = mid + 1;}if (judge)break;}if (!judge)printf("No Solution");return 0;
}

这里要说的是，我们可以通过stl中的binary_search函数，而不用自己写二分查找，这样我们就可以简化代码：

#include<iostream>
#include<string>
#include<algorithm>
using namespace std;
int num[100010] = {
     0 };
int main()
{
    int n, all;scanf("%d %d", &n, &all);for (int i = 0; i < n; i++) {
    int temp;scanf("%d", &num[i]);}sort(num, num + n);bool judge = false;for (int i = 0; num[i] <= all - num[i]; i++) {
    int des = all - num[i];int left = i + 1, right = n;if (binary_search(num + i + 1, num + n, des)) {
    printf("%d %d", num[i], des);judge = true;break;}}if (!judge)printf("No Solution");return 0;
}

Two Pointer方法：

#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<string>
#include<algorithm>
using namespace std;
int num[100010] = {
     0 };
int main()
{
    int n, all;scanf("%d %d", &n, &all);for (int i = 0; i < n; i++) {
    int temp;scanf("%d", &num[i]);}sort(num, num + n);bool judge = false;int i = 0, j = n - 1;while (i < j) {
    if (num[i] + num[j] == all) {
    printf("%d %d", num[i], num[j]);break;}else if (num[i] + num[j] > all) j--;else i++;}if (i >= j) printf("No Solution");return 0;
}

[PAT A1048]Find Coins