1048. Find Coins (25)

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 10⁵ 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.

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

第一个想到的就是二分。之前做过差不多的题。也没多想，直接二分处理

#include<iostream>  #include<cstring>  #include<cstdio>  #include<queue>  #include<stack>  #include<algorithm>  #include<vector> #include<map>using namespace std;int a[100001];int n,m;int f(int s){int zuo=s,you=n-1,fz=n-1;int mid=n-1;while(zuo<you){if(s==mid) break;if(a[s]+a[mid]>m){you=mid;}else if(a[s]+a[mid]<m){zuo=mid+1;}else return a[mid];mid=(zuo+you)/2;}return -1;}int main(){cin>>n>>m;for(int i=0;i<n;i++){scanf("%d",&a[i]);}sort(a,a+n);for(int i=0;i<n;i++){int s=a[i];int num=f(i);    if(num!=-1){    cout<<s<<" "<<num<<endl;    return 0;}}cout<<"No Solution";return 0;}

做完百度看别人的代码。发现其实类似于哥德巴赫猜想。

存在数组里，若一个数存在，对应的数也存在就输出。

欠缺思考。。。。。。。

#include <cstdio>using namespace std;int a[1001];int main() {    int n, m, temp;    scanf("%d %d", &n, &m);    for(int i = 0; i < n; i++) {        scanf("%d", &temp);        a[temp]++;    }    for(int i = 0; i < 1001; i++) {        if(a[i]) {            a[i]--;            if(m > i && a[m-i]) {                printf("%d %d", i, m - i);                return 0;            }            a[i]++;        }    }    printf("No Solution");    return 0;}