HDU2114 Calculate S(n) (取模算术)

Calculate S(n)

Time Limit: 10000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9102    Accepted Submission(s): 3325

Problem Description

Calculate S(n).

S(n)=1³+2³ +3³ +......+n³ .

 

Input

Each line will contain one integer N(1 < n < 1000000000). Process to end of file.

 

Output

For each case, output the last four dights of S(N) in one line.

 

Sample Input

1 2

 

Sample Output

0001 0009

 

Author

天邪

方法一：全算出来。

思路：当n>10000以后，只用关心后四位的数字即可。

知识点：模算术公式：(a+b)%n = ((a%n)+(b%n))%n

　　　　　　　　　　(a-b)%n = ((a%n)-(b%n))%n

　　　　　　　　　　  a*b%n = (a%n)*(b%n)%n

#include <cstdio>
#include <iostream>
#include <cstdlib>
#include <algorithm>
#include <ctime>
#include <cmath>
#include <string>
#include <cstring>
#include <stack>
#include <queue>
#include <list>
#include <vector>
#include <map>
#include <set>
using namespace std;

const int INF=0x3f3f3f3f;
const double eps=1e-10;
const double PI=acos(-1.0);
#define maxn 10005
long long a[maxn];
int main()
{
    long long sum = 0;
    for(int i = 1; i <= 10000; i++)
    {
        sum += (i%10000*i%10000*i%10000)%10000;
        sum  = sum %10000;
        a[i] = sum;
    }
    int n;
    while(~scanf("%d", &n))
    {
        if(n <= 10000)
        {
            printf("%04d
",a[n]);
        }
        else
        {
            int t = n/10000;
            int k = n - t*10000;
            long long sum1 = 0;
            sum1 = ((t%10000)*(a[10000]%10000))%10000 + a[k]%10000;
            printf("%04d
",sum1);

        }
    }
    return 0;
}

方法二：直接公式

#include <stdio.h>
int main()
{
    __int64 n,sum;
    while (scanf("%I64d",&n)!=EOF)
    {
        sum=0;
        n=n%10000;
        sum=(n*(n+1)/2)*(n*(n+1)/2)%10000;// 为了防止计算sum中乘法溢出将n=n%10000;因为要求得后4位数。
        printf("%04I64d
",sum);
    }
    return 0;
}