标签:get http 因数分解 res ble 区间 php 素数 name
被这一道题打崩,打表的题目啊~~~~
类似大区间的素数筛法(POJ2689)
约数和定理:d(n) = (a1 + 1)(a2 + 1)……(an + 1){ai 指的是 质因数分解后质数pi的个数}
代码:
#include <stdio.h> #include <string.h> #include <math.h> #include <iostream> #define input freopen("Multi_4_1003.in", "r", stdin) using namespace std; const int N = 1e6 + 5; typedef long long LL; const int mod = 998244353; int prime[N]; int primecnt; bool vis[N]; LL ans[N]; LL num[N]; void getPrime() { primecnt = 0; for(int i = 2; i <= N; i++) { if(!vis[i]) prime[primecnt++] = i; for(int j = 0; j < primecnt && prime[j] <= N/i; j++) { vis[i*prime[j]] = 1; if(i % prime[j] == 0) break; } } } LL l,r,k; int t; int main() { // input; getPrime(); scanf("%d", &t); while(t--) { scanf("%lld%lld%lld", &l, &r, &k); int m = sqrt(r+0.5); for(LL i = l; i <= r; i++) { ans[i-l] = 1; num[i-l] = i; } for(int i = 0; i < primecnt && prime[i] <= m; i++) { LL p = prime[i]; LL d = l / p + (l % p > 0); if(d == 1) d = 2; for(LL j = d*p; j <= r; j += p) { if(num[j-l] % p == 0) { LL cnt = 0; while(num[j-l] % p == 0){ cnt++; num[j-l] /= p; } cnt = (cnt*k+1) % mod; ans[j-l] = (ans[j-l] * cnt) % mod; } } } LL res = 0; for(LL i = l; i <= r; i++) { if(num[i-l] > 1){ res = (res + ans[i-l]*(k+1))%mod; } else res = (res + ans[i-l])%mod; } printf("%lld\n",res); } }
标签:get http 因数分解 res ble 区间 php 素数 name
原文地址:http://www.cnblogs.com/Alruddy/p/7282601.html