【链接】
【题意】
在这里输入题意
【题解】
考虑删失第i位。
则第i+1位就会代替第i位。
则必定第i+1位比第i位大的话,才对照好。
则从小到大贪心删,找到第一个a[i+1]>a[i]的i.
然后每次删失这样的i就可以了。
【代码】
/* 1.Shoud it use long long ? 2.Have you ever test several sample(at least therr) yourself? 3.Can you promise that the solution is right? At least,the main ideal 4.use the puts("") or putchar() or printf and such things? 5.init the used array or any value? 6.use error MAX_VALUE? 7.use scanf instead of cin/cout? 8.whatch out the detail input require */ /* 必然在这里写完思路再敲代码!!! */ #include <bits/stdc++.h> using namespace std; const int N = 1e5; int n,d,l[N+10],r[N+10]; char s[N+10]; void sc(int x){ int L = l[x],R = r[x]; r[L] = R; l[R] = L; } int main(){ #ifdef LOCAL_DEFINE freopen("rush_in.txt", "r", stdin); #endif ios::sync_with_stdio(0),cin.tie(0); while (cin >> n >> d && n && d){ cin >> (s+1); for (int i = 0;i <= n;i++) l[i] = i-1,r[i] = i+1; int now = r[0]; while (1){ if (d==0)break; int nex = r[now]; while (nex <= n && s[nex]<=s[now]){ now = r[now]; nex = r[nex]; } sc(now); now = l[now]; if (now==0) now = r[now]; d--; } for (int i = r[0];i!=n+1;i=r[i]) cout << s[i]; cout << endl; } return 0; }
，