# A problem is easy

2017/02/09 17:27

## A problem is easy

http://acm.nyist.net/JudgeOnline/problem.php?pid=216

The first line contain a T(T <= 2000) . followed by T lines ,each line contain an integer N (0<=N <= 10^11).

For each case, output the number of ways in one line

2
1
3

0
1

When Teddy was a child , he was always thinking about some simple math problems ,such as “What it’s 1 cup of water plus 1 pile of dough ..” , “100 yuan buy 100 pig” .etc..

One day Teddy met a old man in his dream , in that dream the man whose name was“RuLai” gave Teddy a problem :

Given an N , can you calculate how many ways to write N as i * j + i + j (0 < i <= j) ?

Teddy found the answer when N was less than 10…but if N get bigger , he found it was too difficult for him to solve.
Well , you clever ACMers ,could you help little Teddy to solve this problem and let him have a good dream ?

#include <stdio.h>
#include <math.h>
int main(int argc, char *argv[])
{
int T,count;
long long int n,i,j;
scanf("%d",&T);
while(T--){
count=0;
scanf("%lld",&n);
j=(int)sqrt(n+1)-1;			//计算出i的上限,不能在for循环中重复计算,否则超时
for(i=1;i<=j ;i++){
if( (n+1) % (i+1) == 0)
count++;
}
printf("%d\n",count);
}
return 0;
}


#include<cstring>
#include<cstdio>
#include<map>
#include<string>
#include<algorithm>
#include<vector>
#include<iostream>
#include<cmath>
using namespace std;
#define CLR(arr,val) memset(arr,val,sizeof(arr))

int main()
{
int t,n,cnt=0;
//long long num;
int num;
scanf("%d",&t);
while(t--)
{
//	scanf("%lld",&num);
scanf("%d",&num);
int nn=(int)(sqrt(num+1.0)+0.5);
num++;
cnt=0;
for(int i=2;i<=nn;i++)
if(num%i==0) cnt++;
printf("%d\n",cnt);
}

}


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