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QOJ
ID | 题目 | 提交者 | 结果 | 用时 | 内存 | 语言 | 文件大小 | 提交时间 | 测评时间 |
---|---|---|---|---|---|---|---|---|---|
#94183 | #6137. Sub-cycle Graph | installb# | AC ✓ | 87ms | 4680kb | C++14 | 1.1kb | 2023-04-05 14:48:52 | 2023-04-05 14:48:54 |
Judging History
answer
#include <bits/stdc++.h>
#define rep(i,l,r) for(int i=(l);i<=(r);++i)
#define per(i,r,l) for(int i=(r);i>=(l);--i)
using namespace std;
const int N=1e5+10,P=1e9+7;
int fac[N],ifac[N],f[N],n,m,T;
inline int ksm(int a,int b){
int ans=1;
for(;b;b>>=1,a=1ll*a*a%P) if(b&1) ans=1ll*ans*a%P;
return ans;
}
inline int C(int n,int k){
if(n<k||k<0) return 0;
return 1ll*fac[n]*ifac[k]%P*ifac[n-k]%P;
}
void pre(int n){
f[0]=fac[0]=ifac[0]=1;int tt=(P+1)>>1;
rep(i,1,n) fac[i]=1ll*fac[i-1]*i%P;ifac[n]=ksm(fac[n],P-2);
per(i,n,1) ifac[i-1]=1ll*ifac[i]*i%P;
rep(i,1,n) f[i]=1ll*tt*f[i-1]%P;
}
int main(){
pre(1e5);scanf("%d",&T);
while(T--){
scanf("%d%d",&n,&m);
if(n<m){puts("0");continue;}
if(n==m){printf("%lld\n",1ll*fac[n-1]*ifac[2]%P);continue;}
if(m==0){puts("1");continue;}
int w=n-m,ans=0;
rep(i,0,w){
int p=1ll*f[w-i]%P*C(w,i)%P;
int q=C(m-1,w-i-1);
ans=(ans+1ll*p*q)%P;
}
printf("%lld\n",1ll*ans*ifac[w]%P*fac[n]%P);
}
return 0;
}
详细
Test #1:
score: 100
Accepted
time: 4ms
memory: 4660kb
input:
3 4 2 4 3 5 3
output:
15 12 90
result:
ok 3 number(s): "15 12 90"
Test #2:
score: 0
Accepted
time: 87ms
memory: 4680kb
input:
17446 3 0 3 1 3 2 3 3 4 0 4 1 4 2 4 3 4 4 5 0 5 1 5 2 5 3 5 4 5 5 6 0 6 1 6 2 6 3 6 4 6 5 6 6 7 0 7 1 7 2 7 3 7 4 7 5 7 6 7 7 8 0 8 1 8 2 8 3 8 4 8 5 8 6 8 7 8 8 9 0 9 1 9 2 9 3 9 4 9 5 9 6 9 7 9 8 9 9 10 0 10 1 10 2 10 3 10 4 10 5 10 6 10 7 10 8 10 9 10 10 11 0 11 1 11 2 11 3 11 4 11 5 11 6 11 7 11...
output:
1 3 3 1 1 6 15 12 3 1 10 45 90 60 12 1 15 105 375 630 360 60 1 21 210 1155 3465 5040 2520 360 1 28 378 2940 13545 35280 45360 20160 2520 1 36 630 6552 42525 170100 393120 453600 181440 20160 1 45 990 13230 114345 643545 2286900 4762800 4989600 1814400 181440 1 55 1485 24750 273735 2047815 10239075 3...
result:
ok 17446 numbers