標籤:acm http 分享 set lld mod tor pen deque
題目連結: http://acm.hdu.edu.cn/showproblem.php?pid=1452
題目描述: 讓你求2004^x的所有因數之和, 模29
解題思路: 先將2014質因數分解, 2^2 * 3 * 167, 所以所有因數的個數就是(2x+1)*(x+1)*(x+1) , 我們列出公式, 相當於一個空間直角座標系, 我們先將x, y平面上的點相加, 再加z軸上的, 最後得出公式
ans = (3^(x+1)-1) * (167^(x+1)-1)*(2^(2*x+1)-1)/332 即可, 注意逆元
代碼:
#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;}
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思考: 通過本題自己對逆元的理解也深了一些, 之前一直模稜兩可的.......還有這種題都是有套路的, 題做多了就好了
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
HDU 1452 Happy 2004 數論