C++14(g++5.4) 解法, 执行用时: 18ms, 内存消耗: 476K, 提交时间: 2018-10-10 20:39:50

#include<bits/stdc++.h>
#define ll  long long
//#define int long long
#define inf (1LL<<61)
#define fi first
#define se second
#define pb push_back
#define pa pair<int,int>
#define mkp(a,b) make_pair(a,b)
const int N=1e3+10;
const int mod=998244353;
using namespace std;

double ksm(double a,ll k){double ans=1; while(k){if(k&1) ans=ans*a; a=a*a;k/=2; }return ans; }
double dp[11][N],d[11][N],p[11];
int a[11];
int32_t main()
{
   //ios::sync_with_stdio(0); cin.tie(0);cout.tie(0);
   int t; scanf("%d",&t);
   while(t--)
   {
      int n; scanf("%d",&n);
      for(int i=1;i<=n;i++) scanf("%d%lf",&a[i],&p[i]);
      if(n==1) {puts("1.000000"); continue; }
      for(int i=1;i<=n;i++)
      {
        for(int j=1;j<N;j++)
        {
          dp[i][j]=ksm(1-ksm(p[i],j),a[i]);
          d[i][j]=1-dp[i][j];
        }
      }
      for(int i=1;i<=n;i++)
      {
        double ans=0;
        for(int j=1;j<N-1;j++)
        {
           double tmp=d[i][j]-d[i][j+1];
           for(int k=1;k<=n;k++) (i!=k)&&(tmp*=dp[k][j]);
           ans+=tmp;
        }
        printf("%.6f%c",ans," \n"[i==n]);
      }
   }
return 0;
}
/*

*/

C++11(clang++ 3.9) 解法, 执行用时: 3ms, 内存消耗: 360K, 提交时间: 2018-10-07 18:49:28

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
const int N=100;
double d[11][N],l[11][N];
int n;
double qpow(double a,int b){
	double r=1;
	while(b){
		if(b&1)r=r*a;
		a=a*a;
		b>>=1;
	}
	return r;
}
int main(){
	int T;
	scanf("%d",&T);
	while(T--){
		scanf("%d",&n);
		int a;
		double p,k;
		for(int i=0;i<n;i++){
			scanf("%d%lf",&a,&p);
			k=1;
			for(int j=0;j<N;j++){
				k*=p;
				d[i][j]=qpow(1-k,a);
				l[i][j]=1-d[i][j];
			}
		}
		if(n==1){
			printf("1.000000\n");
			continue;
		}
		for(int i=0;i<n;i++){
			double sum=0;
			for(int j=0;j<N-1;j++){
				double t=l[i][j]-l[i][j+1];
				for(int k=0;k<n;k++)
				if(k!=i)t*=d[k][j];
				sum+=t;
			}
			printf("%lf ",sum);
		}
		printf("\n");
	}
}