标签：center cas desc other can iss stl repr nts
Can you find it? |
| Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/10000 K (Java/Others) |
| Total Submission(s): 1140 Accepted Submission(s): 370 |
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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.
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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.
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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".
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Sample Input
3 3 3 1 2 3 1 2 3 1 2 3 3 1 4 10 |
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Sample Output
Case 1: NO YES NO |
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Author
wangye
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Source
HDU 2007-11 Programming Contest
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/* 二分查找嘛 第一遍想的在三个数组整合到一起查找，但是超内存，想了一下，只整合两个数组的话，时间上虽然复杂了，但是内存小了 */ #include<bits/stdc++.h> #define MAX 505 using namespace std; long long a[MAX],b[MAX],c[MAX],d[MAX*MAX]; int BinarySearch(long long num[],long long end,long long n)/*二分查找*/ { long long l=0,r=end,mid; while(l<=r) { mid=(l+r)/2; if(num[mid]==n) return 1; if(num[mid]>n) r=mid-1; else if(num[mid]<n) l=mid+1; } if(num[l]==n) return 1; return 0; } int main() { //freopen("C:\\Users\\acer\\Desktop\\in.txt","r",stdin); long long L,N,M,t,n; int Case=1; while(scanf("%lld%lld%lld",&L,&N,&M)!=EOF) { for(int i=0;i<L;i++) scanf("%lld",&a[i]); for(int i=0;i<N;i++) scanf("%lld",&b[i]); for(int i=0;i<M;i++) scanf("%lld",&c[i]); long long len=0; for(int i=0;i<L;i++) for(int j=0;j<N;j++) d[len++]=a[i]+b[j]; sort(d,d+len); scanf("%lld",&t); printf("Case %d:\n",Case++); while(t--) { scanf("%lld",&n); int f=0; for(int i=0;i<M;i++) { if(BinarySearch(d,len,n-c[i])) { f=1; break; } } if(f) puts("YES"); else puts("NO"); } } }
标签：center cas desc other can iss stl repr nts
原文地址：http://www.cnblogs.com/wuwangchuxin0924/p/5994656.html
