hdu5358 First One 尺取法 多校联合第六场

First One

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1511    Accepted Submission(s): 457

Problem Description

soda has an integer array a1,a2,…,an. Let S(i,j) be the sum of ai,ai+1,…,aj. Now soda wants to know the value below:

∑i=1n∑j=in(⌊log2S(i,j)⌋+1)×(i+j)

Note: In this problem, you can consider log20 as 0.

 

Input

There are multiple test cases. The first line of input contains an integerT, indicating the number of test cases. For each test case:

The first line contains an integer n(1≤n≤105), the number of integers in the array.
The next line contains n integers a1,a2,…,an(0≤ai≤105).

 

Output

For each test case, output the value.

 

Sample Input

121 1

 

Sample Output

12

  计算上面给的那个公式。

  比赛的时候想了个方法，以每个位置作为终点，二分log每种取值对应的起点，这样的复杂度是O（NlogN*35）左右，但是超时。这道题时间卡的很紧，复杂度O（N*35）的都要运行1000多ms。

  枚举log不同取值的区间范围，在O（N）的复杂度下算出这个范围对应的所有i,j和。采用尺取法，假设以i为起始位置，对应[p1,p2]这个区间为终止位置，这些区间都是满足的，那么当i+1时，一定有终止位置p1'>=p1,p2'>=p2才可能满足。

#include<iostream>#include<cstdio>#include<string>#include<cstring>#include<vector>#include<cmath>#include<queue>#include<stack>#include<map>#include<set>#include<algorithm>using namespace std;typedef long long LL;const LL MAXN=100010;const LL INF=0x3f3f3f3f;const LL SIGMA_SIZE=28;int T,N;int a[MAXN];LL sum[MAXN];//[l,r) i+j的和LL solve(LL l,LL r){    LL ret=0;    int p1=1,p2=1;    for(int i=1;i<=N;i++){        if(p1<i) p1=i;        if(p2<i) p2=i;        while(p1<=N&&sum[p1]-sum[i-1]<l) p1++;        while(p2<=N&&sum[p2]-sum[i-1]<r) p2++;        p2--;        if(p1<=p2&&sum[p1]-sum[i-1]>=l&&sum[p2]-sum[i-1]<r){            LL n=p2-p1+1;            ret+=i*n+p1*n+n*(n-1)/2;        }    }    return ret;}int main(){    freopen("in.txt","r",stdin);    scanf("%d",&T);    while(T--){        scanf("%d",&N);        sum[0]=0;        for(int i=1;i<=N;i++){            scanf("%d",&a[i]);            sum[i]=sum[i-1]+a[i];        }        LL ans=0;        ans+=solve(0,1);        for(int i=0;i<35;i++){            ans+=solve(1LL<<i,1LL<<(i+1))*(i+1);        }        printf("%I64d\n",ans);    }    return 0;}