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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 |
---|---|---|---|---|---|---|---|---|---|
#109295 | #5661. Multi-Ladders | chenshi# | AC ✓ | 1ms | 1756kb | C++ | 799b | 2023-05-28 10:07:24 | 2023-05-28 10:07:28 |
Judging History
answer
#include<cstdio>
using namespace std;
const int MOD=1e9+7;
inline int qp(int b,int f){int res=1;for(;f;f>>=1,b=b*1ll*b%MOD) if(f&1) res=res*1ll*b%MOD;return res;}
int L,n,K,c;
struct Matrix{
int a[2][2];
Matrix operator*(const Matrix&b){
Matrix res;
for(int i=0;i<2;++i) for(int j=0;j<2;++j)
for(int k=res.a[i][j]=0;k<2;++k) res.a[i][j]=(res.a[i][j]+a[i][k]*1ll*b.a[k][j])%MOD;
return res;
}
}A,B;
int main(){
for(scanf("%d",&L);L--;printf("%lld\n",(c*(c-1ll)%MOD*B.a[0][0]+c*(c-1ll)%MOD*(c-2)%MOD*B.a[1][0])%MOD
*qp((c-1+(c-2ll)*(c-2))%MOD,K*(n-1ll)%(MOD-1))%MOD)){
scanf("%d%d%d",&n,&K,&c);
A.a[0][0]=0;A.a[0][1]=c-1;A.a[1][0]=1;A.a[1][1]=c-2;
B.a[0][0]=1;B.a[0][1]=0;B.a[1][0]=0;B.a[1][1]=1;
for(int i=K-2;i;i>>=1,A=A*A) if(i&1) B=B*A;
}
return 0;
}
詳細信息
Test #1:
score: 100
Accepted
time: 1ms
memory: 1500kb
input:
1 2 3 3
output:
162
result:
ok single line: '162'
Test #2:
score: 0
Accepted
time: 1ms
memory: 1756kb
input:
20 2 3 3 1 3 3 10 3 0 10 3 2 1 21 2 1 22 0 2000 15000 2000 12000 30000 200000 1000000000 3 3 2 1000000000 3 2 3 100000000 1000000000 1000000000 10 1000000000 3 100000000 2 1000000000 100000000 1 1000000000 10 1 1000000000 100000000 1 1000 100000000 1000000000 1000000000 0 1000000000 1000000000 1 100...
output:
162 6 0 0 0 0 349400141 243010659 52489881 53690844 176686901 218103365 558243892 991895211 693053429 883715672 80402569 0 0 311752813
result:
ok 20 lines