标签:size amp [1] res 函数 std sigma 使用 .cpp
题意:
f(n) = sigma(gcd(i,n)) 1 <= i <= n
g(n) = sigma(f(d)) d | n
n = x1 * x2 * ... * xm
其中 x[i+1] = (a * x[i] + b) % c + 1
1 <= m <= 10^18
1 <= c <= 10^7
1 <= x[1],a,b <= c
首先,发现f,g函数都是积性函数
并且推下公式:g[n] = n * (k1 + 1 ) * (k2 + 1) * ...
n = p1^k1 * p2^k2 * ....
复杂度O(c) = O(10^7)
公式很容易推,这道题主要是空间太少了,开了数组最后都只能去掉,一个数组使用多次
然后时限也很紧,1300ms,用C++11交了好几发,一直是1400ms左右,改为C++交就1062ms了,
这个要注意
还有个地方,+0LL 被我打成 +1LL,还一直没有发现,wa了好多发。。
代码:
//File Name: nod1643.cpp //Author: long //Mail: 736726758@qq.com //Created Time: 2016年12月22日 星期四 17时15分27秒 #include <bits/stdc++.h> #define LL long long const int MAXN = 10000000 + 1; const int N = 664579 + 1; const int P = (int)1e9 + 7; int fir[MAXN],prime[MAXN],num[MAXN]; LL g,m,x,a,b,c; int tot,C; LL qp(LL x,LL y){ LL res = 1; for(;y>0;y>>=1){ if(y & 1){ res = res * x; if(res >= P) res %= P; } x = x * x; if(x >= P) x %= P; } return res; } void init(){ memset(fir,-1,sizeof(fir)); tot = 0; for(int i=2,j;i<MAXN;++i){ if(fir[i] == -1){ prime[tot++] = i; fir[i] = tot - 1; } for(j=0;j<tot;++j){ if((LL)i * prime[j] >= MAXN) break; fir[i * prime[j]] = j; if(i % prime[j] == 0) break; } } } void cal_num(){ // memset(prime,0,sizeof(prime)); prime[x] = 1,num[x] = 1,fir[1] = x; C = x; g *= x; LL l = 0,r = 0,len = 1; for(LL i=2;i<=m;++i){ x = (a * x + b) % c + 1; if(x > C) C = x; // printf("i = %d x = %lld\n",i,x); if(!prime[x]){ ++num[x]; prime[x] = i; fir[i] = x; g = g * x; if(g >= P) g %= P; } else{ l = prime[x],r = i; len = r - l; break; } } if(r){ int rest = (m - r + 1) % len + l - 1; LL dive = (m - r + 1) / len; LL u = 1,v = 1,w = dive % P,tmp; for(LL i=l,now;i<r;++i){ now = w; v = fir[i]; u = u * v; if(u >= P) u %= P; if(i <= rest){ ++now; g = g * v; if(g >= P) g %= P; } tmp = num[v] + now; if(tmp >= P) tmp -= P; num[v] = tmp; } g = g * qp(u,dive); if(g >= P) g %= P; } } void cal_sum(LL c){ for(LL i=c,id,p,u;i>1;--i){ if(!num[i]) continue; if(prime[fir[i]] == i){ g = g * (num[i] + 1LL); if(g >= P) g %= P; } else{ id = fir[i]; p = prime[id]; u = num[p] + num[i]; if(u >= P) u -= P; num[p] = u; u = num[i / p] + num[i]; if(u >= P) u -= P; num[i / p] = u; } } } void solve(){ g = 1; cal_num(); init(); cal_sum(C); } int main(){ scanf("%lld %lld %lld %lld %lld",&m,&x,&a,&b,&c); solve(); printf("%lld\n",g); return 0; }
标签:size amp [1] res 函数 std sigma 使用 .cpp
原文地址:http://www.cnblogs.com/-maybe/p/6213054.html