[原]NYOJ-216-A problem is easy
大学生程序代写
难度：3
描述
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 ?
输入
The first line contain a T(T <= 2000) . followed by T lines ,each line contain an
integer N (0<=N <= 10^11).
输出
For each case, output the number of ways in one line
样例输入
2
1
3
样例输出
0
1
上传者
苗栋栋*/
/*
双重循环肯定超时的 直接用一个判断条件能减少时间
i*j+i+j =N 经过观察，可以变形为i*j+i+j+1=N+1,也就是说，可以进一步变形为(i+1)*(j+1)=N+1
所以i从2判断到sqrt(n+1)即可
*/
#include<stdio.h>
#include<math.h>//不需要long long int
int main(){
int m,i;
scanf("%d",&m);
while(m--){
int n;
scanf("%d",&n);
{
if((n +1)%i==0)
count++;
}
printf("%d\n",count);
}
return 0;
}
/*A problem is easy
时间限制：1000 ms | 内存限制：65535 KB难度：3
描述
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 ?
输入
The first line contain a T(T <= 2000) . followed by T lines ,each line contain an
integer N (0<=N <= 10^11).
输出
For each case, output the number of ways in one line
样例输入
2
1
3
样例输出
0
1
上传者
苗栋栋*/
/*
双重循环肯定超时的 直接用一个判断条件能减少时间
i*j+i+j =N 经过观察，可以变形为i*j+i+j+1=N+1,也就是说，可以进一步变形为(i+1)*(j+1)=N+1
所以i从2判断到sqrt(n+1)即可
*/
#include<stdio.h>
#include<math.h>//不需要long long int
int main(){
int m,i;
scanf("%d",&m);
while(m--){
int n;
scanf("%d",&n);
int count = 0;
int s=sqrt(n+1);
for(i=2;i<=s;++i){
if((n +1)%i==0)
count++;
}
printf("%d\n",count);
}
return 0;
}
作者：chao1983210400 发表于2013-7-20 12:33:26 原文链接
阅读：12 评论：0 查看评论
