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)=1+2 +3 +......+n .

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-;

const double PI=acos(-1.0);

#define maxn 10005

long long a[maxn];

int main()

{

    long long sum = ;

    for(int i = ; i <= ; i++)

    {

        sum += (i%*i%*i%)%;

        sum  = sum %;

        a[i] = sum;

    }

    int n;

    while(~scanf("%d", &n))

    {

        if(n <= )

        {

            printf("%04d\n",a[n]);

        }

        else

        {

            int t = n/;

            int k = n - t*;

            long long sum1 = ;

            sum1 = ((t%)*(a[]%))% + a[k]%;

            printf("%04d\n",sum1);

        }

    }

    return ;

}

方法二：直接公式

#include <stdio.h>

int main()

{

    __int64 n,sum;

    while (scanf("%I64d",&n)!=EOF)

    {

        sum=;

        n=n%;

        sum=(n*(n+)/)*(n*(n+)/)%;// 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\n",sum);

    }

    return ;

}

模算术

HDU2114 Calculate S(n) (取模算术)

Calculate S(n)