[Codeforces Round #159 (Div. 2)]A. Sockets
2013-01-18 15:22
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地址:http://codeforces.com/contest/257/problem/A
快排(一开始快排还写错了,死循环),然后就是比较所需插孔和已有插孔的个数了
注意别忘了不需要排插的情况,此时答案为0
快排(一开始快排还写错了,死循环),然后就是比较所需插孔和已有插孔的个数了
注意别忘了不需要排插的情况,此时答案为0
#include<stdio.h> int n,m,k; int f[50]; int ones(int l,int r) { int i=l,j=r,flag=0,key=f[l]; int t1; while(i<j) { if(0==flag) { if(f[j]>key) { t1=f[j]; f[j]=f[i]; f[i]=t1; flag=1; } else { j--; } } else { if(f[i]<key) { t1=f[j]; f[j]=f[i]; f[i]=t1; flag=0; } else { i++; } } } return i; } void quick(int l,int r) { int mid; if(l<r) { mid=ones(l,r); quick(l,mid-1); quick(mid+1,r); } } int main() { int i,ans=-1; scanf("%d %d %d",&n,&m,&k); for(i=0;i<n;i++) { scanf("%d",&f[i]); } quick(0,n-1); for(i=0;i<n;i++) { if(0==i && k>=m) {ans=0;break;} if(f[i]+k-1<m) k=k+f[i]-1; else {ans=i+1;break;} } printf("%d\n",ans); return 0; }
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