HDU 4342——History repeat itself—————

HDU 4342——History repeat itself——————【数学规律】

History repeat itself

Time Limit: 1000ms

Memory Limit: 32768KB

This problem will be judged on HDU. Original ID: 4342
64-bit integer IO format: %I64d Java class name: Main

Type:

Tom took the Discrete Mathematics course in the 2011,but his bad attendance angered Professor Lee who is in charge of the course. Therefore, Professor Lee decided to let Tom face a hard probability problem, and announced that if he fail to slove the problem there would be no way for Tom to pass the final exam.
As a result , Tom passed.
History repeat itself. You, the bad boy, also angered the Professor Lee when September Ends. You have to faced the problem too.
The problem comes that You must find the N-th positive non-square number M and printed it. And that's for normal bad student, such as Tom. But the real bad student has to calculate the formula below.

So, that you can really understand WHAT A BAD STUDENT YOU ARE!!

Input

There is a number (T)in the first line , tell you the number of test cases below. For the next T lines, there is just one number on the each line which tell you the N of the case.
To simplified the problem , The N will be within 231 and more then 0.

Output

For each test case, print the N-th non square number and the result of the formula.

Sample Input

Sample Output

Source

2012 Multi-University Training Contest 5

解题思路1：找规律。

#include<bits/stdc++.h>
using namespace std;
typedef __int64 INT;
INT dfs(INT c){
    INT cc=(INT)sqrt(c);
    if(cc==0){
        return 0;
    }
    INT ret=(c-cc*cc+1)*cc;
    return ret+dfs(cc*cc-1);
}
int main(){
    int t;
    INT n,m;
    scanf("%d",&t);
    while(t--){
        scanf("%I64d",&n);
        INT nn=(INT)sqrt(n);
        m=n+nn;
        INT mm=(INT)sqrt(m);
        if(mm>nn)    //这条件开始写错了
            m++;
        INT ans=dfs(m);
        printf("%I64d %I64d
",m,ans);
    }
    return 0;
}

思路2：数学推导。http://www.cnblogs.com/kuangbin/archive/2012/08/08/2628794.html

#include<bits/stdc++.h>
using namespace std;
typedef __int64 INT;
INT dfs(INT c){
    INT cc=(INT)sqrt(c);
    if(cc==0){
        return 0;
    }
    INT ret=(c-cc*cc+1)*cc;
    return ret+dfs(cc*cc-1);
}
int main(){
    int t;
    INT n,m;
    scanf("%d",&t);
    while(t--){
        scanf("%I64d",&n);
        INT mm=(INT)ceil((1+sqrt(1+4*n))/2)-1;
        m=mm+n;
        INT ans=dfs(m);
        printf("%I64d %I64d
",m,ans);
    }
    return 0;
}