10056 - What is the Probability ?
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描述:求第I个人赢的概率,公式为:p*(1-p)^(I-1)*(1+(1-p)^n+(1-p)^2n(1-p)^3n....),由此可以得到化简公式p*(1-p)^(I-1)*((1-(1-p)^n^2)/(1-(1-p)^n))(等比数列),由于此项(1-(1-p)^n^2)接近于1,就当作1来处理,所以可得p*(1-p)^(I-1)/(1-(1-p)^n),但是p为零的时候要特殊处理#include <cstdio>#include <cmath>int main(){ // freopen("in.txt","r",stdin); int t,n,q; double p; scanf("%d",&t); while(t--) { scanf("%d %lf %d",&n,&p,&q); if(p<0.0000001) printf("0.0000\n"); else printf("%.4lf\n",p*pow(1-p,q-1)/(1-pow(1-p,n)) ); } return 0;}
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