「2017 山东一轮集训 Day6」子序列（矩阵快速幂）

/*
找出了一个dp式子
是否能够倍增优化 
我推的矩阵不太一样
是
1 0 0 0 0
0 0 0 0 -1 
0 0 1 0 0
0 0 0 1 0
0 1 0 0 2

求得逆矩阵大概就是

1 0 0 0 0
0 2 0 0 1
0 0 1 0 0
0 0 0 1 0
0 -1 0 0 0 
*/
#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<queue>
#define ll long long 
#define M 100010
#define log lllgggi 
#define mmp make_pair
using namespace std;
int read()
{
    int nm = 0, f = 1;
    char c = getchar();
    for(; !isdigit(c); c = getchar()) if(c == '-') f = -1;
    for(; isdigit(c); c = getchar()) nm = nm * 10 + c - '0';
    return nm * f;
}
const int mod = 1000000007;
char s[M];

void add(int &x, int y)
{
    x += y;
    x -= x >= mod ? mod : 0;
    x += x < 0 ? mod : 0;
}
struct Mx{
    int a[10][10];
    Mx()
    {
        memset(a, 0, sizeof(a));
    }
}be[9], iv[9], an[M], bn[M];


Mx mul(Mx a, Mx b)
{
    Mx c;
    for(int i = 0; i <= 9; i++)
    {
        for(int j = 0; j <= 9; j++)
        {
            for(int k = 0; k <= 9; k++)
            {
                add(c.a[i][k], 1ll * a.a[i][j] * b.a[j][k] % mod);
            }
        }
    }
    return c;
}



int n, a[M], sum, q, f[10], g[10]; 

int main()
{
    scanf("%s", s + 1);
    n = strlen(s + 1);
    for(int i = 1; i <= n; i++) a[i] = s[i] - 'a';
    for(int k = 0; k <= 8; k++)
    {
        for(int i = 0; i <= 8; i++)
        {
            if(i == k)
            {
                be[k].a[9][k] = 1;
                be[k].a[k][9] = mod - 1;
                iv[k].a[i][i] = 2;
                iv[k].a[i][9] = 1;
                iv[k].a[9][i] = mod - 1;
            }
            else
            {
                be[k].a[i][i] = 1;
                iv[k].a[i][i] = 1;
            }
        }
        be[k].a[9][9] = 2;
    }
    for(int i = 0; i <= 9; i++) an[0].a[i][i] = bn[0].a[i][i] = 1;
    for(int i = 1; i <= n; i++) an[i] = mul(an[i - 1], be[a[i]]), bn[i] = mul(iv[a[i]], bn[i - 1]);
    q = read();
    
    while(q--)
    {
        int l = read(), r = read();
        memset(f, 0, sizeof(f));
        f[9] = 1;
        memset(g, 0, sizeof(g)); 
        for(int i = 0; i <= 9; i++)
        {
            for(int j = 0; j <= 9; j++)
            {
                add(g[i], 1ll * f[j] * bn[l - 1].a[j][i] % mod);
            }
        }
        memcpy(f, g, sizeof(f));
        int ans = 0;
        for(int j = 0; j <= 9; j++)
        {
            add(ans, 1ll * f[j] * an[r].a[j][9] % mod);
        }
        cout << (ans - 1 + mod) % mod << "\n";
    }
    return 0;
}