hdu 1452 Happy 2004 找循环节

题目链接：hdu 1452 Happy 2004 

题意：求2004^X 约数之和S与29取模的结果。

分析：2004 = 2 * 2 * 3 * 167， 2004^x = 2^(2x) * 3^x * 167^x， 求约数之和S ，我们可以构造多项式：

S=(2^0+2^1+...+2^(2x)) * (3^0+3^1+...+3^x) * (167^0+167^1+...+167^x). 

打印出X∈[1,100] 范围内的结果。找出循环节。周期为28。

#include <cmath>#include <queue>#include <vector>#include <cstdio>#include <string>#include <cstring>#include <iomanip>#include <iostream>#include <algorithm>using namespace std;//#pragma comment(linker, "/STACK:1024000000,1024000000")#define FIN             freopen("input.txt","r",stdin)#define FOUT            freopen("output.txt","w",stdout)#define fst             first#define snd             second#define lson            l, mid, rt << 1#define rson            mid + 1, r, rt << 1 | 1typedef __int64  LL;//typedef long long LL;typedef unsigned int uint;typedef pair<int, int> PII;typedef pair<LL, LL> PLL;const int MAXN = 100000 + 5;const int INF = 0x3f3f3f3f;const int MOD = 29;int X, S;int v[] = {2, 3, 167};int d[3];int res[MOD + 5];int main() {#ifndef ONLINE_JUDGE    FIN;    // FOUT;#endif // ONLINE_JUDGE    for(X = 1; X < MOD; X++) {        d[0] = (X << 1);        d[1] = d[2] = X;        S = 1;        int SUM;        SUM = 0;        for(int i = 0, a = 1; i <= d[0]; i++, a = (a * v[0]) % MOD) {            SUM = (SUM + a) % MOD;        }        S = (S * SUM) % MOD;        SUM = 0;        for(int i = 0, b = 1; i <= d[1]; i++, b = (b * v[1]) % MOD) {            SUM = (SUM + b) % MOD;        }        S = (S * SUM) % MOD;        SUM = 0;        for(int i = 0, c = 1; i <= d[2]; i++, c = (c * v[2]) % MOD) {            SUM = (SUM + c) % MOD;        }        S = (S * SUM) % MOD;        res[X] = S;    }    while(~scanf("%d", &X) && X) {        printf("%d\n", res[(X - 1) % 28 + 1]);    }    return 0;}