google code jam 2014 RC_A
2014-06-24 15:45
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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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