标签:acm http 分享 set lld mod tor pen deque
题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=1452
题目描述: 让你求2004^x的所有因数之和, 模29
解题思路: 先将2014质因数分解, 2^2 * 3 * 167, 所以所有因数的个数就是(2x+1)*(x+1)*(x+1) , 我们列出公式, 相当于一个空间直角坐标系, 我们先将x, y平面上的点相加, 再加z轴上的, 最后得出公式
ans = (3^(x+1)-1) * (167^(x+1)-1)*(2^(2*x+1)-1)/332 即可, 注意逆元
代码:
#include <iostream> #include <cstdio> #include <string> #include <vector> #include <cstring> #include <iterator> #include <cmath> #include <algorithm> #include <stack> #include <deque> #include <map> #define lson l, m, rt<<1 #define rson m+1, r, rt<<1|1 #define mem0(a) memset(a,0,sizeof(a)) #define sca(x) scanf("%d",&x) #define de printf("=======\n") typedef long long ll; using namespace std; const int mod = 29; ll q_power( ll a, ll b ) { ll ret = 1; while( b ) { if( b & 1 ) ret = ret * a % mod; b >>= 1; a = a * a % mod; } return ret % mod; } ll exgcd(ll a, ll b, ll &x, ll &y) { if(b == 0) { x = 1; y = 0; return a; } else { ll ret = exgcd(b, a%b, x, y); ll tmp = x; x = y; y = tmp - a / b * y; return ret; } } ll inv(ll a) { ll x, y; exgcd(a, mod, x, y); return (x % mod + mod) % mod; } int main() { ll x; while( scanf( "%lld", &x ) == 1 && x ) { ll ans; ll a = q_power(3, x+1); ll b = q_power(167, x+1); ll c = q_power(2, 2*x+1); ans = (a*b*c-a*c-b*c+c-a*b+a+b-1) * inv(332) % mod; printf( "%lld\n", ans ); } return 0; }
思考: 通过本题自己对逆元的理解也深了一些, 之前一直模棱两可的.......还有这种题都是有套路的, 题做多了就好了
a*b*c-a*c-b*c+c-a*b+a+b
a*b*c-a*c-b*c+c-a*b+a+b
a*b*c-a*c-b*c+c-a*b+a+b
标签:acm http 分享 set lld mod tor pen deque
原文地址:http://www.cnblogs.com/FriskyPuppy/p/7447741.html