HDU - 2141 Can you find it? 二分
2018-02-11 22:07
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Can you find it?
Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/10000 K (Java/Others)Total Submission(s): 35824 Accepted Submission(s): 8840
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 Input3 3 3 1 2 3 1 2 3 1 2 3 3 1 4 10
Sample OutputCase 1: NO YES NO
Author wangye
Source HDU 2007-11 Programming Contest
题目大意:先输入三列数字,分别标记为Ai,Bi,Ci,判断是否存在Ai,Bj,Ck使得其相加为后面所输入的数。思路:先将前两列的数相加,用一个数组储存,然后使用sort将其排序,最后二分查找。
#include<stdio.h> #include<algorithm> using namespace std; int L_[505], N_[505], M_[505]; int sum_[250005]; int main() { int L, N, M; int S; int x,num=0,temp; while (~scanf("%d %d %d", &L, &N, &M)) { num++; int k = 0; for(int i = 0; i < L; i++) { scanf("%d", &L_[i]); } for (int i = 0; i < N; i++) { scanf("%d", &N_[i]); } for (int i = 0; i < M; i++) { scanf("%d", &M_[i]); } for (int i = 0; i < N; i++) for (int j = 0; j < M; j++) sum_[k++] = N_[i] + M_[j]; sort(sum_, sum_ + k); scanf("%d", &S); printf("Case %d:\n", num); while (S--) { int flag = 0; scanf("%d", &x); for (int i = 0; i < L; i++) { temp = x - L_[i]; int left = 0,right=k-1; int mid; while (left <= right) { mid = (left + right) / 2; if (sum_[mid] > temp) { right=mid-1; } else if (sum_[mid]<temp) { left=mid+1; } else { flag = 1; break; } } if (flag==1) { printf("YES\n"); break; } } if (flag == 0) { printf("NO\n"); } } } return 0; }
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