1048. Find Coins (25)

Eva loves to collect coins from all over the universe, including some other planets like Mars. One day she visited auniversal shopping mall which could accept all kinds of coins as payments. However, there was a special requirement ofthe payment: for each bill, she could only use exactly two coins to pay the exact amount. Since she has as many as 10⁵coins 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.

Input Specification:

Each input file contains one test case. For each case, the first line contains 2 positive numbers: N (<=10⁵, the total number of coins) and M(<=10³, 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.

Output Specification:

For each test case, print in one line the two face values V₁ and V₂ (separated by a space) such that V₁ + V₂ = M and V₁ <= V₂. If such a solution is not unique, output the one with the smallest V₁. If there is no solution, output "No Solution" instead.

Sample Input 1:

8 151 2 8 7 2 4 11 15

Sample Output 1:

4 11

Sample Input 2:

7 141 8 7 2 4 11 15

Sample Output 2:

No Solution

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提交代码

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这么简单的题目，肯定是卡你时间或者占用的空间。。。。

先暴力一下。

#include <iostream>#include <vector>#include <algorithm>using namespace std;/* run this program using the console pauser or add your own getch, system("pause") or input loop */int N,M;int main(int argc, char** argv) {int i=0;scanf("%d %d",&N,&M);vector<int> coins(N);for(i=0; i<N; i++){scanf("%d",&coins[i]);}sort(coins.begin(),coins.end());bool flag=false;for(i=0; i<coins.size() && coins[i]<=M; i++){int v1=coins[i];int tmp=M-v1;for(int j=i+1; j<coins.size() && coins[j] <=tmp; j++){int v2=coins[j];if(v2!=tmp){continue;}else{flag=true;printf("%d %d\n",v1,v2);break;}}if(flag) break;}if(!flag) printf("No Solution\n");return 0;}

然后换一个思路，存储钱的个数，这样只有N的耗时。

AC

#include <iostream>#include <vector>#include <algorithm>using namespace std;/* run this program using the console pauser or add your own getch, system("pause") or input loop */#define MAX 1010int N,M;int main(int argc, char** argv) {int i=0;scanf("%d %d",&N,&M);vector<int> coins(MAX);coins.assign(MAX,0);for(i=0; i<N; i++){int tmp;scanf("%d",&tmp);coins[tmp]++;}bool flag=false;for(i=1; i<=M/2; i++){coins[i]--;coins[M-i]--;if(coins[i]>=0 && coins[M-i]>=0){printf("%d %d",i,M-i);flag=true;break;}}if(!flag) printf("No Solution\n");return 0;}