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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 |
---|---|---|---|---|---|---|---|---|---|
#144262 | #6137. Sub-cycle Graph | sihan_88 | AC ✓ | 89ms | 3916kb | C++14 | 1005b | 2023-08-21 16:04:34 | 2023-08-21 16:04:37 |
Judging History
answer
#include<cstdio>
typedef long long ll;
const int MOD=int(1e9)+7;
const int INV2=(MOD+1)>>1;
inline ll ksm(ll b,int p)
{
ll res=1;
for(;p;p>>=1,b=b*b%MOD) if(p&1) (res*=b)%=MOD;
return res;
}
const int N=100010;
ll fact[N],fact_inv[N],P[N];
inline ll C(int n,int m)
{
if(n<0||m<0||n<m) return 0;
return fact[n]*fact_inv[m]%MOD*fact_inv[n-m]%MOD;
}
inline ll F(int n,int m)
{
//a_i\ge2 \sum_{a_i}=n
if(!m) return n==0;
return C(n-m-1,m-1);
}
int main()
{
fact[0]=P[0]=1;
for(int i=1;i<N;++i) fact[i]=fact[i-1]*i%MOD,P[i]=P[i-1]*INV2%MOD;
fact_inv[N-1]=ksm(fact[N-1],MOD-2);
for(int i=N-2;i>=0;--i) fact_inv[i]=fact_inv[i+1]*(i+1)%MOD;
int T,n;
ll m;
scanf("%d",&T);
while(T--)
{
scanf("%d%lld",&n,&m);
if(m>n) puts("0");
else if(m==n) printf("%lld\n",fact[n-1]*INV2%MOD);
else
{
ll sum=0;
for(int i=0;i<=n-m;++i) sum=(sum+C(n-m,i)*P[i]%MOD*F(m+i,i))%MOD;
printf("%lld\n",sum*fact[n]%MOD*fact_inv[n-m]%MOD);
}
}
return 0;
}
詳細信息
Test #1:
score: 100
Accepted
time: 2ms
memory: 3884kb
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: 89ms
memory: 3916kb
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