Codeforces Round #420 (Div. 2) E. Okabe and El Psy Kongroo [矩阵快速幂]

题目无聊到求多次快速幂…

#pragma comment(linker, "/STACK:102400000,102400000")#include<cstdio>#include<algorithm>#include<cmath>#include<cstring>#include<vector>#include<map>#include<string>#include<iostream>using namespace std;using namespace std;typedef long long ll;const ll mod=1000000007;int n;ll m;struct Matrix{    ll mp[20][20];    Matrix()    {        memset(mp,0,sizeof(mp));    }};Matrix mul(Matrix a,Matrix b){    int i,j,k;    Matrix c;    for(i=0; i<n; i++)    {        for(j=0; j<n; j++)        {            c.mp[i][j]=0;            for(k=0; k<n; k++)            {                c.mp[i][j]=(c.mp[i][j]+a.mp[i][k]*b.mp[k][j]%mod)%mod;            }        }    }    return c;}Matrix mypow(Matrix t,ll x){    Matrix c;    for(int i=0; i<n; i++)    c.mp[i][i]=1;    while(x)    {        if(x&1)            c=mul(c,t);        t=mul(t,t);        x>>=1;    }    return c;}long long a[1024],b[1024],c[1024];int main(){    int t;    long long k;    n=20;    scanf("%d %lld",&t,&k);    for(int i=0;i<t;i++)    {       scanf("%lld%lld%lld",a+i,b+i,c+i);    }    Matrix A;    A.mp[0][0]=1;    for(int p=0;p<t;p++)    {        Matrix B;        for(int i=0;i<=c[p];i++)        {            if(i==0)            {                B.mp[0][0]=B.mp[1][0]=1;            }            else                 //if(i<=c[p])            {                B.mp[i-1][i]=B.mp[i][i]=B.mp[i+1][i]=1;            }        }        //if(a[p]==k) continue;        //A =mul(A,B);        //if(min(k,b[p])-a[p]-1<=0) continue;        if(p)         for(int i=min(c[p-1],c[p])+1;i<n;i++)             A.mp[0][i]=0;         B=mypow(B,min(k,b[p])-a[p]);        A=mul(A,B);    }    printf("%lld\n",A.mp[0][0]);    return 0;}