ACM: 限时训练题解-Runtime Error-二分查找

1. Runtime Error

Bahosain was trying to solve this simple problem, but he got a Runtime Error on one of the test cases, can you help him by solving it?

Given an array of N non-negative integers and an integer K, your task is to find two integers X and Y from the given array such that X × Y = K.

The chosen numbers must have different indices in the   array.

Input

The first line of input contains T (1 ≤ T ≤ 128), the number of test   cases.

The first line of each test case contains two integers: N (2 ≤ N ≤ 100,000) and K (1 ≤ K ≤ 100,000). The next line contains N space-separated integers, each between 0 and   100,000.

Output

For each test case, if there is no solution, print -1 on a single line. Otherwise print a single line with two space-separated integers X Y (X ≤   Y), where X and Y are two numbers from the given array and X × Y = K.

If there is more than one possible solution, print the one with the minimum   X.

/*
题意：
从给出的N个数中找出两个数，乘积为 K；枚举x 二分搜索 y
*/ #include"iostream"
#include"algorithm"
#include"cstdio"
#include"cstring"
#include"cmath"
#define MX 100000 + 50
using namespace std;int a[MX];int main() {
int T,k,n;
scanf("%d",&T);
while(T--) {
scanf("%d%d",&n,&k);
for(int i=0; i<n; i++) {
scanf("%d",&a[i]);
}
sort(a,a+n);
int ans1=0,ans=0;
for(int i=0; i<n-1; i++) {
ans1=i;int l=i+1,r=n-1,mid;
while(l<=r) {
mid=(r+l)/2;
if(a[i]*a[mid]>k) {
r=mid-1;
} else if(a[i]*a[mid]<k) {
l=mid+1;
} else if(a[i]*a[mid]==k){
ans=mid;
break;
}
}
if(ans)break;}
if(ans)
printf("%d %d\n",a[ans1],a[ans]);
else printf("-1\n");
}
return 0;
}