于神之怒加强版
Time Limit: 80 Sec Memory Limit: 512 MBSubmit: 1184 Solved: 535
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Description
给下N,M,K.求
Input
输入有多组数据,输入数据的第一行两个正整数T,K,代表有T组数据,K的意义如上所示,下面第二行到第T+1行,每行为两个正整数N,M,其意义如上式所示。
Output
如题
Sample Input
1 2
3 3
3 3
Sample Output
20
HINT
1<=N,M,K<=5000000,1<=T<=2000
题解:JudgeOnline/upload/201603/4407.rar
Source
命题人:成都七中张耀楠,鸣谢excited上传。
1 #include<bits/stdc++.h> 2 #pragma GCC optimize(2) 3 #pragma G++ optimize(2) 4 #include<iostream> 5 #include<algorithm> 6 #include<cmath> 7 #include<cstdio> 8 #include<cstring> 9 10 #define ll long long 11 #define inf 1000000000 12 #define mod 1000000007 13 #define N 5000007 14 using namespace std; 15 inline int read() 16 { 17 int x=0,f=1;char ch=getchar(); 18 while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();} 19 while(isdigit(ch)){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();} 20 return x*f; 21 } 22 23 int F[N],f[N],flag[N],k,tot,p[N],ans; 24 inline int gpow(int x,int y) 25 { 26 int ans=1; 27 while (y) 28 { 29 if (y&1) ans=(ll)ans*x%mod; 30 y>>=1;x=(ll)x*x%mod; 31 } 32 return ans; 33 } 34 void preparation() 35 { 36 F[1]=1; 37 for (int i=2;i<N;i++) 38 { 39 if (!flag[i]){f[i]=gpow(i,k);F[i]=f[i]-1;p[++tot]=i;} 40 for (int j=1;j<=tot&&i*p[j]<N;j++) 41 { 42 flag[i*p[j]]=1; 43 if (i%p[j])F[i*p[j]]=(ll)F[i]*F[p[j]]%mod; 44 else{F[i*p[j]]=(ll)F[i]*f[p[j]]%mod;break;} 45 } 46 } 47 for (int i=1;i<N;i++) (F[i]+=F[i-1])%=mod; 48 } 49 int main() 50 { 51 int Case=read();k=read(); 52 preparation(); 53 while (Case--) 54 { 55 int n=read(),m=read();if (n>m) swap(n,m);ans=0; 56 for (int i=1,pos=0;i<=n;i=pos+1) 57 { 58 pos=min(n/(n/i),m/(m/i)); 59 (ans+=1LL*(n/i)*(m/i)%mod*(F[pos]-F[i-1])%mod)%=mod; 60 } 61 printf("%d\n",(ans+mod)%mod); 62 } 63 return 0; 64 }