Ternary String
時間限制：C/C++ 4秒，其他語言8秒
空間限制：C/C++ 131072K，其他語言262144K
64bit IO Format: %lld
題目描述
A ternary string is a sequence of digits, where each digit is either 0, 1, or 2.
Chiaki has a ternary string s which can self-reproduce. Every second, a digit 0 is inserted after every 1 in the string, and then a digit 1 is inserted after every 2 in the string, and finally the first character will disappear.
For example, 212'' will become

11021” after one second, and become “01002110” after another second.
Chiaki would like to know the number of seconds needed until the string become an empty string. As the answer could be very large, she only needs the answer modulo (109 + 7).
輸入描述:
There are multiple test cases. The first line of input is an integer T indicates the number of test cases. For each test case:
The first line contains a ternary string s (1 ≤ |s| ≤ 105).
It is guaranteed that the sum of all |s| does not exceed 2 x 106.
輸出描述:
For each test case, output an integer denoting the answer. If the string never becomes empty, output -1 instead.
示例1
輸入
複製
3
000
012
22
輸出
複製
3
93
45

操作分析：
1、遇到0，直接刪除，操作次數+1
2、遇到1，考慮之前已經操作了x次，那麼這個1後面已經多生出了x個0，這個時候需要在經過x + 2次操作才能刪完
3、遇到2，考慮之前已經操作了x次，然後打個表可以發現，之後還需要經過3 * （2^（x + 1）- 1 ） - x 次操作才能全部刪完。

可見計算a*2 ^ x mod (10 ^ 9 + 7),於是也得計算x mod phi(10 ^ 9 + 7),考慮到x可能也是之前某些2^x的組合，因此還得算x mod phi(phi(10 ^ 9 + 7)),以此類推
另外考慮到若干次後，mod一定是一個2的倍數，可以提前算好答案，來優化。
嗨，主要是推公式很難推出來！！！！

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long LL;
const int inf = 0x3f3f3f3f;
const int N = 100005;
char s[N];
LL phi[35]={1000000007,1000000006,500000002,//3
243900800,79872000,19660800,5242880,2097152,//5
1048576,524288,262144,131072,65536,32768,16384,//7
8192,4096,2048,1024,512,256,128,64,32,16,8,4,2,1,1};//13
LL pow_mod(LL n,LL m,LL p)
{
    LL x = n;
    LL sum = 1;
    while(m)
    {
        if(m & 1)
            sum = (sum * x) % p;
        m >>= 1;
        x = (x * x) % p;
    }
    return sum;
}
LL dfs(int pos,int num)
{
    if(num > 27) return 0;
    if(pos == -1) return 0;
    if(s[pos] == '0'){
        return (dfs(pos - 1,num) + 1LL) % phi[num];
    }
    else if(s[pos] == '1'){
        return (dfs(pos - 1,num) * 2LL + 2LL) % phi[num];
    }else{
        //後面加+phi[num]) % phi[num]是為了防止出現負數 
        return (((3LL * pow_mod(2,dfs(pos - 1,num + 1) + 1,phi[num])) - 3LL) % phi[num] + phi[num]) % phi[num];
    }
}
int main()
{
    int t;
    while(~scanf("%d",&t))
    {
        while(t--)
        {
            scanf("%s",s);
            int len = strlen(s);
            printf("%lld\n",dfs(len - 1,0));
        }
    }
    return 0;
}