GalaxyOJ-987 (Linear Screen + Thinking)

topic

Rounded statistics

analysis

• For the first six test points, you can quickly sift out prime numbers with a linear sieve, and then calculate each point with the number theorem. For k, you can actually multiply cnt by k every time *cnt.
• Then the next four big data points will not work. If you can't filter out so many prime numbers, you won't be able to judge one by one by force.
• Think about it and discover that r-l<=10^6, and the prime number after r^2 can only appear once to the power or 0 times.
• So you can screen out sqrt(r)prime numbers within the put l~rall the numbers except violence, get that done sqrt(r)after less than a few, if the l~rnumber is still not 1, then they must be greater than sqrt(r)the number of times a certain quality, it directly multiplied by the value of d k+1can be.

program

#include <cstdio>
#define Ha 998244353
typedef long long ll;
ll l,r,k,N,d[1000005];
ll p[1000005],rest[1000005],num,ans;
bool f[1000005];

int main(){
    scanf("%lld%lld%lld",&l,&r,&k); l--;
    N=r-l;
    for (ll i=l+1; i<=r; i++) d[i-l]=1,rest[i-l]=i;
    for (ll x=2; x*x<=r; x++){
        if (!f[x]){
            p[++num]=x;
            ll ytx=l/x*x+x,cnt;
            for (cnt=0; ytx<=r; ytx+=x,cnt=0){
                while (rest[ytx-l]%x==0)
                    rest[ytx-l]/=x,cnt++;
                cnt=(cnt*k+1)%Ha;
                d[ytx-l]=d[ytx-l]*cnt%Ha;
            }
        }
        for (ll i=1; i<=num && x*p[i]<=1000000; i++) f[x*p[i]]=1;
    }
    for (ll i=l+1; i<=r; i++) if (rest[i-l]>1) d[i-l]=d[i-l]*(k+1)%Ha;
    for (ll i=l+1; i<=r; i++) ans=(ans+d[i-l])%Ha;
    printf("%lld",ans);
}