google code jam 2014 RC_A

来源：互联网发布：海关数据编辑：程序博客网时间：2024/05/23 00:09

1 三分普通序列，使其中子序列和最大的最小。思想：二分。时间n*log（n).

#include <cstdio>#include <iostream>using namespace std;int T,N,p,q,r,s;long long sum[1000001];long long ABS(long long a){return a>0?a:-a;}long long MAX(long long a,long long b){return a>b?a:b;}int main(){freopen("A-large-practice.in","r",stdin);freopen("out","w",stdout);cin>>T;for(int cas=1; cas<=T; ++cas){cin>>N>>p>>q>>r>>s;int tmp=q%r;sum[0]=tmp+s;for(int i=1; i<N; ++i){tmp+=p; tmp%=r;sum[i]=sum[i-1]+tmp+s;}long long best=0x7fffffffffffffffLL;for(int i=0; i<N; ++i){long long left=(i-1<0?0:sum[i-1]);long long right=sum[N-1]-(i-1<0?0:sum[i-1]);int l=i,r=N-1,m=(l+r)/2;while(l<r){m=(l+r)/2;long long left2=sum[m]-(i-1<0?0:sum[i-1]);long long right2=sum[N-1]-sum[m];if(left2<right2){l=m+1;}else{r=m;}}int tm[3]={m-1,m,m+1};if(cas==60){int tmp=3;}for(int j=0; j<3; ++j){int m=tm[j];if(m<i||m>=N) continue;long long tmp01=MAX(left,MAX(sum[m]-(i-1<0?0:sum[i-1]),sum[N-1]-sum[m]));if(best>tmp01){best=tmp01;}}}//printf("%lld\n",best);printf("Case #%d: %.10f\n",cas,1-1.0*best/sum[N-1]);}}

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