标签：图片 SM max tle 需要 for enter 分享图片 cep
Problem DescriptionNow you are given one non-negative integer n in 10-base notation, it will only contain digits (‘0‘-‘9‘). You are allowed to choose 2 integers i and j, such that: i!=j, 1≤i<j≤|n|, here |n| means the length of n’s 10-base notation. Then we can swap n[i] and n[j].
For example, n=9012, we choose i=1, j=3, then we swap n[1] and n[3], then we get 1092, which is smaller than the original n.
Now you are allowed to operate at most M times, so what is the smallest number you can get after the operation(s)?
Please note that in this problem, leading zero is not allowed!
InputThe first line of the input contains an integer T (T≤100), indicating the number of test cases.
Then T cases, for any case, only 2 integers n and M (0≤n<10^1000, 0≤M≤100) in a single line.
Output Sample Input Sample Output#include <iostream> #include<algorithm> #include<vector> #include<cstring> #include<queue> #include<string> using namespace std; #define N_MAX 10000+10 #define V_MAX 1000+10 #define INF 0x3f3f3f3f int m; string n; void change(string &s,int n){//当前调整s的第n位 char c=‘9‘+1;int id; for(int i=s.size()-1;i>n;i--){ if(c>s[i]){ if(n==0&&s[i]==‘0‘)continue; c=s[i];id=i; } } if(c<s[n])swap(s[n],s[id]); } int main(){ int t;scanf("%d",&t); while(t--){ cin>>n>>m; int cnt=0; string s; while(m){ s=n; while(s==n&&cnt<s.size()-1){ change(n,cnt); cnt++; } if(cnt>=s.size()-2)break;//已经不需要交换了 m--; } cout<<n<<endl; } return 0; }
foj 2111 Problem 2111 Min Number
标签：图片 SM max tle 需要 for enter 分享图片 cep
原文地址：https://www.cnblogs.com/ZefengYao/p/8822896.html
