Moonfang's Birthday&&2011 Multi-University Training Contest 3 - Host by BIT
2012-01-07 11:59
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题意:有一个人要过生日了,为了给他过生日他的同伴给他对钱,,但是每个同伴的钱都不一样,但是为了尽可能的平均,让你设计一种算法,,求出在尽可能平均的情况下每人对的钱数。。
思路:先把同伴拥有的钱数从小到大排序,,,每次取平均数和拥有的钱数中少的那个为该同伴最后需要对的钱数。。。。
AC代码:
思路:先把同伴拥有的钱数从小到大排序,,,每次取平均数和拥有的钱数中少的那个为该同伴最后需要对的钱数。。。。
AC代码:
#include<iostream> #define N 10001 #include<algorithm> #include<string.h> using namespace std; typedef struct { int val; int id; int w; }Node; Node s ; bool cmp1(const Node& a,const Node& b) { if(a.val!=b.val) return a.val<b.val; else return a.id>b.id; } bool cmp(const Node& a,const Node& b) {return a.id<b.id;} int main() { int Case; cin>>Case; while(Case--) { int n,m; cin>>n>>m; int sum=0; for(int i=0;i!=m;++i) { cin>>s[i].val; sum+=s[i].val; s[i].id=i; } if(sum<n) {cout<<"IMPOSSIBLE"<<endl;continue;} sort(s,s+m,cmp1); sum=n; for(int i=0;i<m;++i) { s[i].w=min(s[i].val,sum/(m-i)); sum-=s[i].w; } sort(s,s+m,cmp); for(int i=0;i<m-1;++i) cout<<s[i].w<<" "; cout<<s[m-1].w<<endl; }return 0; }
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