Can you find it? （二分）

2019-11-06 08:06:14

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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. InputThere 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. OutputFor 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 331410Sample OutputCase 1:NOYESNO
n三次方的复杂度会超时。
三个数组，把第一个和第二个数组组合起来，构成sum数组保存所有和的可能（有重复无所谓），然后再在sum数组中
二分根据 sum[i]+c[j]=X，在sum数组二分X-c[j]的值，一旦某一个j找到一个对应的sum[i]，就YES。时间复杂度下降
为n^2+nlogn
#include <iostream>#include <cstdio>#include <cmath>#include <queue>#include <stack>#include <map>#include <algorithm>#include <vector>#include <string>#include <cstring>#include <sstream>#define INF 100000000using namespace std;int L,N,M,S;int a[505];int b[505];int c[505];int sum[250005];int k;int x;int tmp;int main(){    int kase=1;    while(scanf("%d%d%d",&L,&N,&M)==3)    {        for(int i=0;i<L;i++)        {            scanf("%d",&a[i]);        }        for(int i=0;i<N;i++)        {            scanf("%d",&b[i]);        }        for(int i=0;i<M;i++)        {            scanf("%d",&c[i]);        }        k=0;        for(int i=0;i<L;i++)        {            for(int j=0;j<N;j++)            {                sum[k++]=a[i]+b[j];            }        }        sort(sum,sum+k);        scanf("%d",&S);        printf("Case %d:/n",kase++);        while(S--)        {            scanf("%d",&x);            int flag=0;            for(int j=0;j<M;j++)            {                tmp=x-c[j];                int left=0,right=k-1;                int mid;                while(left<=right)                {                    mid=(left+right)/2;                    if(sum[mid]>tmp)                    {                        right=mid-1;                    }                    else if(sum[mid]<tmp)                    {                        left=mid+1;                    }                    else                    {                        flag=1;                        break;                    }                }                if(sum[left]==tmp&&left<k || sum[right]==tmp&&right>=0)                {                    flag=1;                }                if(flag==1)                {                    printf("YES/n");                    break;                }            }            if(flag==0)            {                printf("NO/n");            }        }    }    return 0;}

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