PAT Advanced Level 1079
2014-03-27 02:01
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题目地址:http://pat.zju.edu.cn/contests/pat-a-practise/1079
代码如下:
代码如下:
#include <stdio.h> #include <math.h> #include <vector> using namespace std; vector<int> s[100011]; vector<int> ss[100011]; int main() { int n; double up,prp; scanf("%d%lf%lf",&n,&up,&prp); for(int i=0;i<n;i++){ s[i].clear(); int t; scanf("%d",&t); s[i].push_back(t); if(t==0){ int tt; scanf("%d",&tt); s[i].push_back(tt); } else{ for(int j=0;j<t;j++){ int tmp; scanf("%d",&tmp); s[i].push_back(tmp); } } } double res=0; int layer=0; ss[0].clear(); ss[0].push_back(0); while(ss[layer].size()>0){ ss[layer+1].clear(); for(int i=0;i<(signed)ss[layer].size();i++){ int ttt=ss[layer][i]; if(s[ttt][0]==0){ res+=pow((1+prp/100),layer)*up*s[ttt][1];//使用pow函数能节省不少时间 } else{ for(int j=1;j<=s[ttt][0];j++){ ss[layer+1].push_back(s[ttt][j]); } } } layer++; } printf("%.1lf\n",res); return 0; }
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