Happy 2004&&http://acm.hdu.edu.cn/showproblem.php?pid=1452

Total Submission(s): 609 Accepted Submission(s): 431


Problem Description
Consider a positive integer X,and let S be the sum of all positive integer divisors of 2004^X. Your job is to determine S modulo 29 (the rest of the division of S by 29).

Take X = 1 for an example. The positive integer divisors of 2004^1 are 1, 2, 3, 4, 6, 12, 167, 334, 501, 668, 1002 and 2004. Therefore S = 4704 and S modulo 29 is equal to 6.

Input
The input consists of several test cases. Each test case contains a line with the integer X (1 <= X <= 10000000).

A test case of X = 0 indicates the end of input, and should not be processed.

Output
For each test case, in a separate line, please output the result of S modulo 29.

Sample Input
   
   
   
   
1 10000 0

Sample Output
   
   
   
   
6 10
题意:关于求一个数所有因子之和的问题,一种是等比数列求和,一种是乘法逆元。对于这一题我只能说我SB了,明知道hdu不支持long long类型(如果使用就会超时),却视而不见,导致对自己的算法产生了怀疑,疯狂调试了近两个小时,还是木有找出错误,最后蓦然发现我却用了long long 类型,着实伤不起啊!!有木有,,,,
法一:
#include<iostream>
#include<string.h>
#include<string>
#define N  29
#include<cstdio>
using namespace std;
int a[][3]={2,3,167,2,1,1};
typedef __int64 L;
L pow(L a,L b)
{
    L ans=1;
    while(b)
    {
        if(1&b) ans=(ans*a)%N;
            a=(a%N*a%N)%N;
            b=b>>1;
    }
    return ans;
}
L _pow(L a,L b)
{
    if(a==0) return  0;
    else if(a==1||b==0) return 1;
    else {
         if(1&b) return (_pow(a,b/2)%N)*((1+pow(a,b/2+1))%N);
         else   return ((_pow(a,b/2-1)%N)*((1+pow(a,b/2+1))%N)+pow(a,b/2))%N;
    }
}
int main()
{
    L x;
    while(~scanf("%I64d",&x),x)
    {
        L ans=1;
        for(int i=0;i!=3;++i)
            ans=(ans%N*(_pow(a[0][i],a[1][i]*x)%N))%N;
        printf("%I64d\n",ans);
    }return 0;
}

法二:
*#include<cstdio>
#include<string.h>
#include<iostream>
#define N 29
using namespace std;
typedef __int64  L;
L _pow(L a,L b)
{
	L ans=1;
	while(b)
	{
		if(1&b) ans=(ans*a)%N;
		    a=(a%N*a%N)%N;
			b=b>>1;
	}
	return ans;
}
int main()
{
	L x;
	while(~scanf("%I64d",&x),x)
	{
		L a=_pow(2,(x*2+1));
		L b=_pow(3,(x+1));
		L c=_pow(22,(x+1));
		L d=((a-1)*(b-1)*15*(c-1)*18)%N;
		printf("%I64d\n",d);
	}return 0;
(a+k*29)^29%29=a%29;


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