/*
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;
}
转载于:https://www.cnblogs.com/luoyibujue/p/10619055.html
