d(x)表示x的约数个数,让你求(l,r<=10^12,r-l<=10^6,k<=10^7)
#include<cstdio> using namespace std; #define MOD 998244353ll #define MAXP 1000100 typedef long long ll; ll x,y; int T,K; bool isNotPrime[MAXP+10]; int num_prime,prime[MAXP+10]; void shai() { for(long i = 2 ; i < MAXP ; i ++) { if(! isNotPrime[i]) prime[num_prime ++]=i; for(long j = 0 ; j < num_prime && i * prime[j] < MAXP ; j ++) { isNotPrime[i * prime[j]] = 1; if( !(i % prime[j])) break; } } } ll b[1000010],a[1000010]; int main(){ scanf("%d",&T); shai(); for(;T;--T){ scanf("%lld%lld%d",&x,&y,&K); for(ll i=x;i<=y;++i){ a[i-x+1ll]=i; b[i-x+1ll]=1; } for(int i=0;i<num_prime;++i){ ll t=x/(ll)prime[i]*(ll)prime[i]+(ll)(x%(ll)prime[i]!=0)*(ll)prime[i]; for(ll j=t;j<=y;j+=(ll)prime[i]){ int cnt=0; while(a[j-x+1ll]%(ll)prime[i]==0){ a[j-x+1ll]/=(ll)prime[i]; ++cnt; } b[j-x+1ll]=(b[j-x+1ll]*(((ll)cnt*(ll)K%MOD+1ll)%MOD))%MOD; } } ll ans=0; for(ll i=x;i<=y;++i){ if((a[i-x+1ll]>1ll)){ b[i-x+1ll]=(b[i-x+1ll]*((ll)K+1ll))%MOD; } ans=(ans+b[i-x+1ll])%MOD; } printf("%lld ",ans); } return 0; }