紫书 例题10-6 无关的元素 UVa1635

题意：给定n个数a1,a2····an,依次求出相邻两个数值和，将得到一个新数列，重复上述操作，最后结果将变为一个数，问这个数除以m的余数与那些数无关？例如n=3,m=2时，第一次得到a1+a2,a2+a3，在求和得a1+2*a2+a3，它除以2的余数和a2无关。1<=n<=10^5, 2<=m<=10^9

分析：

通过试验可以发现，本题等价于求解C(n-1,i)的组合数中有哪些是m的倍数，可以利用唯一分解定理来判断：事先分解m，随后利用递推式计算每一项中包含m的素因数的指数即可。

代码如下：

////Created by BLUEBUFF 2016/1/11//Copyright (c) 2016 BLUEBUFF.All Rights Reserved//#pragma comment(linker,"/STACK:102400000,102400000")//#include <ext/pb_ds/assoc_container.hpp>//#include <ext/pb_ds/tree_policy.hpp>//#include <ext/pb_ds/hash_policy.hpp>//#include <bits/stdc++.h>#include <set>#include <map>#include <queue>#include <stack>#include <cmath>#include <cstdio>#include <time.h>#include <cstdlib>#include <cstring>#include <complex>#include <sstream> //isstringstream#include <iostream>#include <algorithm>using namespace std;//using namespace __gnu_pbds;typedef long long LL;typedef unsigned long long uLL;typedef pair<int, LL> pp;#define REP1(i, a, b) for(int i = a; i < b; i++)#define REP2(i, a, b) for(int i = a; i <= b; i++)#define REP3(i, a, b) for(int i = a; i >= b; i--)#define CLR(a, b)     memset(a, b, sizeof(a))#define MP(x, y)      make_pair(x,y)template <class T1, class T2>inline void getmax(T1 &a, T2 b) { if (b>a)a = b; }template <class T1, class T2>inline void getmin(T1 &a, T2 b) { if (b<a)a = b; }const int maxn = 100010;const int maxm = 1e5+5;const int maxs = 10;const int maxp = 1e3 + 10;const int INF  = 1e9;const int UNF  = -1e9;const int mod  = 1e9 + 7;const int rev = (mod + 1) >> 1; // FWT//const double PI = acos(-1);//headint fac1[100][2]; //fac[i][0]存放素因数， fac[i][1]存放其指数int fac2[100], a[maxn];void factor(int m){    int &num = fac1[0][0];//fac[0][0]是表头，存放总的个数，用引用比较方便    num = 0;    for(int i = 2; i * i <= m; i++){        if(m % i == 0){            fac1[++num][0] = i;            fac1[num][1] = 0;            do{                fac1[num][1]++;                m /= i;            }while(m % i == 0); //将i除干净        }    }    if(m > 1)//如果分解到最后m仍然大于1,说明它是一个素数。注意：如果只是判断素因子有哪些，可以没有此处判断，否则必须有此步    {        fac1[++num][0] = m;        fac1[num][1] = 1;    }}bool check(int n, int j)//按照递推公式来计算第j项，检查唯一分解式的指数{    int num = fac1[0][0];    int a = n - j, b = j;    for(int i = 1; i <= num; i++){        int p = fac1[i][0];        int &q = fac2[i];        for(; a % p == 0; a /= p, q++);//为了提高效率，只用检验m的分解式中的素因数即可        for(; b % p == 0; b /= p, q--);    }    for(int i = 1; i <= num; i++){        if(fac1[i][1] > fac2[i]){            return false;        }    }    return true;}int main(){    int n, m;    while(scanf("%d%d", &n, &m) != EOF)    {        int cnt = 0;        factor(m);        memset(fac2, 0, sizeof(fac2));        for(int i = 1; i < n; i++){            if(check(n, i)){                a[cnt++] = i + 1;            }        }        printf("%d\n", cnt);        for(int i = 0; i < cnt; i++){            printf("%d%c", a[i], i == (cnt - 1) ? '\n' : ' ');        }    }    return 0;}