Untitled
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 774 Accepted Submission(s): 423
Problem Description
There is an integer a and n integers b1,…,bn. After selecting some numbers from b1,…,bn in any order, say c1,…,cr, we want to make sure that a mod c1 mod c2 mod… mod cr=0 (i.e., a will become the remainder divided by ci each time, and at the end, we want a to become 0). Please determine the minimum value of r. If the goal cannot be achieved, print −1 instead.
Input
The first line contains one integer T≤5, which represents the number of testcases.
For each testcase, there are two lines:
1. The first line contains two integers n and a (1≤n≤20,1≤a≤106).
2. The second line contains n integers b1,…,bn (∀1≤i≤n,1≤bi≤106).
For each testcase, there are two lines:
1. The first line contains two integers n and a (1≤n≤20,1≤a≤106).
2. The second line contains n integers b1,…,bn (∀1≤i≤n,1≤bi≤106).
Output
Print T answers in T lines.
Sample Input
2 2 9 2 7 2 9 6 7
Sample Output
2 -1
不知道为什么必须判断比他大的数啊,老是超时,判断了就不超时了。
#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; int t ; int n,a; int x[25]; bool cmp(int a,int b){ return a >b; } int dfs(int u,int depth){ if(u == 0)return depth; int minn = n+1; for(int i = depth;i<n;i++){ int z = u>=x[i]?dfs(u % x[i],depth+1):n+1; if( z < minn)minn = z; } return minn; } int main(){ int t; scanf("%d",&t); while(t--){ scanf("%d%d",&n,&a); for(int i = 0;i<n;i++){ scanf("%d",&x[i]); } sort(x,x+n,cmp); int minn = n+1; for(int i = 0;i<n;i++){ int z = a >= x[i]?dfs(a % x[i],1):n+1; if( z < minn) minn = z; } printf("%d\n",minn == n+1?-1:minn); } }