USACO-Section1.1 Greedy Gift Givers
2017-05-25 22:11
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Greedy Gift Givers
本体只需按照所给输入按顺序操作即可解。为求方便用map做了姓名与编号间的映射。
代码如下:
/* ID: xhzdcyy1 PROB: gift1 LANG: C++ */ #include <iostream> #include <fstream> #include <cstring> #include <map> #define cin fin #define cout fout using namespace std; int main() { ofstream fout ("gift1.out"); ifstream fin ("gift1.in"); int NP,NG[10]={0}; string name[10],tmp; map<string,int> m; cin>>NP; for(int i=0;i<NP;i++){ cin>>tmp; name[i]=tmp; m[tmp]=i; } for(int i=0;i<NP;i++){ cin>>tmp; int t1=m[tmp],money,c; cin>>money>>c; for(int j=0;j<c;j++){ cin>>tmp; NG[t1]-=(money/c); int t2=m[tmp]; NG[t2]+=(money/c); } } for(int i=0;i<NP;i++){ cout<<name[i]<<" "<<NG[i]<<endl; } return 0; }
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