bzoj:4407: Shinzhi Fury Enhanced Edition

Description to the next n,m,k.

Input has more than one set of data, the first line of input data two positive integer t,k, representing a T group of data, the meaning of K as shown above, the second line below to the t+1 line, each behavior two positive integer n,m, the meaning of the above-mentioned figure.

Output title

Sample Input1 2
3 3Sample Output -How does it feel to be a topic of water? mention a d^k at the back of Mo, and then put μ and d^k together, pre-treatment out on the good. but pay attention to the pre-processing when writing Nlogn notation will T, to be placed in the Linear sieve O (n) sieve out, 30+s. (Maya, what is 10+s?

#include <cstdio>#include<algorithm>#defineN 5000000using namespacestd;Const intmod=1e9+7;intt,n,m,k,k,num=0, p[n],mu[n+1],ne,f[n+1],mmh;BOOLbo[n+1];inlineintMintx) { while(X>=mod) X-=mod; while(x<0) X+=mod;returnx;} InlineintMiintAintx) {MMH=1;  while(x) {if(x&1) mmh=1ll*mmh*a%MOD; X>>=1; A=1ll*a*a%MOD; }    returnMmh;} InlineintMinintAintb) {returnA<b?a:b;}intMain () {registerinti,j,k,l; scanf ("%d%d",&t,&j); mu[1]=1; f[1]=1;  for(i=2; i<=n;i++){        if(!bo[i]) p[++num]=i,mu[i]=-1, F[i]=mi (i,k)-1;  for(j=1;j<=num&& (ne=p[j]*i) <=n;j++) {Bo[ne]=1; if(I%p[j]) mu[ne]=-mu[i],f[ne]=1ll*f[i]*f[p[j]]%mod;Else{Mu[ne]=0; F[ne]=1ll*f[i]*mi (p[j],k)%MOD;  Break; }        }    }     for(i=1; i<=n;i++) F[i]=m (f[i]+f[i-1]);  while(t--) {scanf ("%d%d",&n,&m); if(n>m) swap (N,M);  for(i=1, j=0, mmh=0; i<=n;j=i++) I=min (n/(n/i), m/(m/i)), Mmh=m (mmh+1ll* (n/i) * (m/i)%mod*m (F[i]-f[j])%MOD); printf ("%d\n", MMH); }}

64280 KB 31400 ms C++/edit 1092 B

bzoj:4407: Shinzhi Fury Enhanced Edition