HDU2141：Can you find it?（二分 + 优化）

Can you find it?

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 27166    Accepted Submission(s): 6840

Problem Description

Give you three sequences of numbers A, B, C, then we give you a number X. Now you need to calculate if you can find the three numbers Ai, Bj, Ck, which satisfy the formula Ai+Bj+Ck = X.

 

Input

There are many cases. Every data case is described as followed: In the first line there are three integers L, N, M, in the second line there are L integers represent the sequence A, in the third line there are N integers represent the sequences B, in the forth line there are M integers represent the sequence C. In the fifth line there is an integer S represents there are S integers X to be calculated. 1<=L, N, M<=500, 1<=S<=1000. all the integers are 32-integers.

 

Output

For each case, firstly you have to print the case number as the form "Case d:", then for the S queries, you calculate if the formula can be satisfied or not. If satisfied, you print "YES", otherwise print "NO".

 

Sample Input

3 3 31 2 31 2 31 2 331410

 

Sample Output

Case 1:NOYESNO

 

Author

wangye

 

Source

HDU 2007-11 Programming Contest

题意：判断能否从三组数据各选一个和为Si。

思路：将前两组的和的情况列举出来，排序后二分第三组即可。

# include <stdio.h># include <algorithm>using namespace std;int main(){    int cas=1, tmp, s, a[3][501];    long long b[250001];    while(~scanf("%d%d%d",&a[0][0],&a[1][0],&a[2][0]))    {        for(int i=0; i<3; ++i)            for(int j=1; j<=a[i][0]; ++j)                scanf("%d",&a[i][j]);        int l, r, mid, cnt = 0;        scanf("%d",&s);        printf("Case %d:\n",cas++);        for(int i=1; i<=a[0][0]; ++i)            for(int j=1; j<=a[1][0]; ++j)                b[cnt++] = a[0][i] + a[1][j];        sort(a[2]+1, a[2]+a[2][0]+1);        sort(b, b+cnt);        cnt = unique(b, b+cnt)-b;//去重优化        while(s--)        {            bool flag = false;            scanf("%d",&tmp);            if(tmp>a[2][a[2][0]]+b[cnt-1] || tmp<a[2][1]+b[0])//优化            {                puts("NO");                continue;            }            for(int i=1; i<=a[2][0]; ++i)            {                l = 0;                r = cnt-1;                mid = (l+r)>>1;                while(l<r)                {                    if(b[mid]+a[2][i]<tmp)                        l = mid+1;                    else                        r = mid;                    mid = (l+r)>>1;                }                if(b[r]+a[2][i]==tmp)                {                    flag = true;                    break;                }            }            if(flag)                puts("YES");            else                puts("NO");        }    }    return 0;}