hdu 2141 Can you find it? 二分
2014-09-04 14:36
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传送门:http://acm.hdu.edu.cn/showproblem.php?pid=2141
Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 11577 Accepted Submission(s): 3031
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
Sample Output
Author
wangye
三个数能否相加得到X; 把头两个相加,存入set. 然后X减去第三个数,一个个减过来. 然后用find在set 中二分来找,找到即是可以相加得到X;
之前全开__int64 MLE 了
开了两个set来分别存 结果TLE了
最后这样都化成int用一个set终于过了.
Can you find it?
Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/10000 K (Java/Others)Total Submission(s): 11577 Accepted Submission(s): 3031
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 3 1 2 3 1 2 3 1 2 3 3 1 4 10
Sample Output
Case 1: NO YES NO
Author
wangye
三个数能否相加得到X; 把头两个相加,存入set. 然后X减去第三个数,一个个减过来. 然后用find在set 中二分来找,找到即是可以相加得到X;
之前全开__int64 MLE 了
开了两个set来分别存 结果TLE了
最后这样都化成int用一个set终于过了.
#include<stdio.h> #include<set> using namespace std; #define ll __int64 int a[600]; int c[600]; int main() { int i,l,m,r,flag,op,tem,j; set<int>my; int cas=1; while(scanf("%d%d%d",&l,&m,&r)!=EOF) { my.clear(); for(i=0;i<l;i++) { scanf("%d",&a[i]); } for(i=0;i<m;i++) { scanf("%d",&tem); for(j=0;j<l;j++) { if((ll)tem+a[j]<=2147483647) my.insert(tem+a[j]); } } for(i=0;i<r;i++) { scanf("%d",&c[i]); } scanf("%d",&op); printf("Case %d:\n",cas++); while(op--) { scanf("%d",&tem); flag=0; for(i=0;i<r;i++) { if(my.find(tem-c[i])!=my.end()) { flag=1; break; } } if(flag) puts("YES"); else puts("NO"); } } return 0; } //2147483647 /* 1 2 1 2147483647 1 0 0 3 */
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