http://cdn.ac.nbutoj.com/Problem/view.xhtml?id=1181

问题描述

The game LOL is an interesting game. Recently we are playing this game. Once I used a champion named Kog'Maw - The Mouth of the Abyss.
For I have to support teammate, I only have a little time to do the last hit to kill the enemy's minions.
There're N (2 <= N <= 10000) minions, each minion has a threat value TVi (1 <= TVi <= 9).
And there's a formula to calculate the total threat value:

I only can kill K (1 <= K < N) minions.
When I kill one minion, the minions after it will move to front for one step. It means when I kill Minion[5], then Minion[6] will be Minion[5] and Minion[7] will be Minion[6] and so on.
How to kill the minions that I can leave the minimum total threat value?

输入

This problem contains several cases.
Each case contains two numbers. The first number T is the total threat value at the very beginning (1 <= T < 10^10001). Then follows an integer K, means the number that minions I can kill.

输出

For each case, you should output the minimum total threat value after I kill K minions.

样例输入

123456 43212311 4

样例输出

12111

法一：采用贪心策略，每次删除单调递减的第一个字符。

AC代码：

#include<iostream>#include<string>using namespace std;int main(){string s;int n;while(cin>>s>>n){ for(int i=0;i<n;++i){for(int j=0;j<s.size();++j)if(j==s.size()-1||s[j]>s[j+1]){s.erase(s.begin()+j);break;}}cout<<s<<endl;}return 0;}

法二：每次在规定范围内删除最小的字符

#include<iostream>#include<string.h>#include<string>using namespace std;string work(string s,int n){int m=s.size()-n;int first=0,last=n;string s1="";while(m--){  int minx=9999999,p;for(int i=first;i<=last;++i)if(s[i]<minx) {p=i;minx=s[i];}    s1+=minx;   first=p+1,last++;}return s1;}int main(){string s;int n;while(cin>>s>>n){cout<<work(s,n)<<endl;}return 0;}