hdu 6069用两次筛法

题意：

求

(∑i=lrd(ik))mod998244353

思路：

PS：两次筛法，第一次筛出所有的素数，第二次用素数去筛区间[L,R]之间的数，然后求出他们对应的值，当处理完之后f[i]>1的话说明是素数。

#include<bits/stdc++.h>using namespace std;typedef long long ll;typedef pair<int,int> P;typedef set<int>::iterator ITER;#define fi first#define se second#define INF 0x3f3f3f3f#define clr(x,y) memset(x,y,sizeof x)const int maxn = 1e6 + 10;const int Mod = 998244353;const int N = 2;struct Matrix{Matrix(){clr(m,0);}int m[N][N];};Matrix Mul(Matrix a,Matrix b){Matrix c;for(int i = 0; i < N; i ++){for(int k = 0; k < N; k ++){if(a.m[i][k] == 0)continue;for(int j = 0; j < N; j ++)c.m[i][j] += a.m[i][k] * b.m[k][j] % Mod,c.m[i][j] %= Mod;}}return c;}ll Mul(ll a,ll b){a %= Mod;b %= Mod;ll ret = 0;while(b){if(b & 1){ret += a;if(ret > Mod)ret -= Mod;}b >>= 1;a = (a << 1) % Mod;}return ret;}Matrix pows(Matrix x,ll n){Matrix ret;for(int i = 0; i < 2; i ++)ret.m[i][i] = 1;while(n){if(n & 1)ret = Mul(ret,x);x = Mul(x,x);n >>= 1;}return ret;}ll pows(ll x,ll n){ll ret = 1;while(n){if(n & 1)ret = ret * x % Mod;x = x * x % Mod;n >>= 1;}return ret;}ll powss(ll x,ll n){ll ret = 1;while(n){if(n & 1)ret = Mul(ret,x);x = Mul(x,x);n >>= 1;}return ret;}ll gcd(ll x,ll y){return y ? gcd(y,x % y):x;}ll lcm(ll x,ll y){return x /gcd(x,y) * y;}ll euler(int n){int ret = n;for(int i = 2; i * i<= n; i ++)if(n % i == 0){ret = ret / i * (i - 1);while(n % i == 0)n /= i;}if(n > 1)ret = ret / n * (n - 1);return ret;}void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){d = a;x = 1;y = 0;return;}ex_gcd(b,a % b,d ,y,x);y -=a/b * x;}inline int lowbit(int x){return x &(-x);}//void update(int x,int val){for(int i = x; x < maxn; i += lowbit(i))tree[i] += val;}//void get(int x){int ret = 0;for(int i = x; i > 0 ;i -= lowbit(i))ret += tree[i];return ret;}//int finds(int x){return fa[x] == x ? x : (fa[x] = finds(fa[x]));}bool is_[maxn];ll prime[maxn];int len;ll f[maxn],g[maxn];ll l,r,k;void Init(){len = 0;clr(is_,true);for(int i = 2; i < maxn; i ++){if(is_[i]){prime[len ++] = i;for(int j = i * 2; j < maxn; j += i)is_[j] = false;}}}void fun(int p){for(ll i = l/p * p; i <= r; i += p)if(i >= l){//    cout << p << " -> " << i << endl;int cnt = 0;while(f[i - l] % p == 0)cnt ++,f[i - l] /= p;g[i - l] = g[i - l] * (cnt * k + 1) % Mod;}}int main(){Init();int Tcase;scanf("%d",&Tcase);for(int ii = 1; ii <= Tcase; ii ++){scanf("%I64d%I64d%I64d",&l,&r,&k);ll n = r - l;for(int i = 0; i <= n; i ++)f[i] = i + l,g[i] = 1;for(int i = 0; i < len;i ++){if(prime[i] * prime[i] > r)break;fun(prime[i]);}ll ans = 0;for(int i = 0; i <= n; i ++){if(f[i] > 1)g[i] = g[i] * (k + 1) % Mod;//cout << i + l<< " " << g[i] << endl;ans = (ans + g[i]) % Mod;}printf("%I64d\n",ans);}return 0;}