2017 Multi-University Training Contest
来源:互联网 发布:施耐德可编程编程软件 编辑:程序博客网 时间:2024/06/08 13:20
题目:http://acm.split.hdu.edu.cn/showproblem.php?pid=6050
对于任意i>=1,当j>=3时,有
上式的推导:
1.可以打表找规律。。。
2.类似归纳法推
之后就是官方题解:
http://bestcoder.hdu.edu.cn/blog/2017-multi-university-training-contest-2-solutions-by-%E7%94%B5%E5%AD%90%E7%A7%91%E6%8A%80%E5%A4%A7%E5%AD%A6/点击打开链接
两次矩阵快速幂可得
#include<bits/stdc++.h>using namespace std;long long b[2][2][2]={-1,0,0,-1, 1,-2,-1,0};const int mo=1e9+7;struct ma{ long long a[2][2];}A;ma mult(ma &a,ma &b){ ma c; memset(c.a,0,sizeof(c.a)); for(int i=0;i<2;i++) { for(int j=0;j<2;j++) { for(int k=0;k<2;k++) { c.a[i][j]=(c.a[i][j]+a.a[i][k]*b.a[k][j])%mo; } } } return c;}ma pow_ma(ma t,long long x){ ma ans; int i,j; for(i=0;i<2;i++) ans.a[i][i]=1; ans.a[0][1]=ans.a[1][0]=0; while(x) { if(x&1) ans=mult(ans,t); t=mult(t,t); x>>=1; } return ans;}int main(){ int T; cin>>T; long long n,m; A.a[0][0]=0,A.a[0][1]=2,A.a[1][0]=1,A.a[1][1]=1; ma ans; while(T--) { scanf("%lld %lld",&n,&m); ans=pow_ma(A,n); for(int i=0;i<2;i++) { for(int j=0;j<2;j++) ans.a[i][j]=(ans.a[i][j]+b[n&1][i][j]+mo)%mo; } ans=pow_ma(ans,m-1); printf("%lld\n",(ans.a[0][0]+ans.a[1][0])%mo); } return 0;}
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- 2017 Multi-University Training Contest
- 2017 Multi-University Training Contest
- 2017 Multi-University Training Contest
- 2017 Multi-University Training Contest
- 2017 Multi-University Training Contest
- 2017 Multi-University Training Contest
- 2017 Multi-University Training Contest
- 2017 Multi-University Training Contest
- 2017 Multi-University Training Contest
- 2017 Multi-University Training Contest
- 2017 Multi-University Training Contest
- 2017 Multi-University Training Contest
- #2017 Multi-University Training Contest
- 2017 Multi-University Training Contest
- #2017 Multi-University Training Contest
- 2017 Multi-University Training Contest
- 2017 Multi-University Training Contest
- 2017 Multi-University Training Contest
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