Can you find it?
Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 3988 Accepted Submission(s): 997
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.
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".
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
在做此题的时候,如果单纯的用多重循环来实现会超时,正确的思路是先合并数组+排序+二分查找。下面是我的代码:
#include <stdio.h> #include <stdlib.h> int cmp(const void *a,const void *b) { return (*(int *)a-*(int *)b); } int main() { int f(int a[500*500],int x,int y,int z); int i,j,m,n,s,t,x,y,z,k; int a[500],b[500*500],c[500],d[1000]; z=1; while(scanf("%d %d %d",&n,&m,&t)!=EOF) { y=0; for(i=0;i<=n-1;i++) { scanf("%d",&a[i]); } for(i=0;i<=m-1;i++) { scanf("%d",&x); for(j=0;j<=n-1;j++) { b[y]=a[j]+x; y+=1; } } for(i=0;i<=t-1;i++) { scanf("%d",&c[i]); } scanf("%d",&x); for(i=0;i<=x-1;i++) { scanf("%d",&d[i]); } qsort(b,n*m,sizeof(b[0]),cmp); printf("Case %d:\n",z); for(i=0;i<=x-1;i++) { for(j=0;j<=t-1;j++) { s=d[i]-c[j]; k=f(b,0,n*m-1,s); if(k==1) { break; } } if(j==t) { printf("NO\n"); }else { printf("YES\n"); } } z+=1; } return 0; } int f(int a[500*500],int x,int y,int z) { int k,mid; k=0; while(x<=y) { mid=(x+y)/2; if(a[mid]==z) { k=1; break; }else if(a[mid]>z) { y=mid-1; }else { x=mid+1; } } return (k); }