FZU 2108(dfs模拟，大数取余)

 K

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status Practice FZU 2108

Description

Given one non-negative integer A and one positive integer B, it’s very easy for us to calculate A Mod B. Here A Mod B means the remainder of the answer after A is divided by B. For example, 7 Mod 5 = 2, 12 Mod 3 = 0, 0 Mod 3 = 0.

 

In this problem, we use the following rules to express A.

(1) One non-empty string that only contains {0,1,2,3,4,5,6,7,8,9} is valid.

For example, 123, 000213, 99213. (Leading zeros is OK in this problem)

(2) If w is valid, then [w]x if valid. Here x is one integer that 0<x<10.

For example, [012]2=012012, [35]3[7]1=3535357.

(3) If w and v are valid, then wv is valid.

For example, w=[231]2 and v=1, then wv=[231]21 is valid (which is 2312311).

 

Now you are given A and B. Here A is express as the rules above and B is simply one integer, you are expected to output the A Mod B.

Input

The first line of the input contains an integer T(T≤10), indicating the number of test cases.

Then T cases, for any case, only two lines.

The first line is one non-empty and valid string that expresses A, the length of the string is no more than 1,000.

The second line is one integer B(0<B<2,000,000,000).

You may assume that the length of number A in decimal notation will less than 2^63.

Output

For each test case, output A Mod B in a single line.

Sample Input

3

[0]9[[1]2]3

10007

[[213231414343214231]5]1

10007

[0012]1

1

Sample Output

1034 3943 0

题解：【】后面代表里面的数字出现的次数；直接暴力会超内存

比如说[123]2=123123=123*1000+123。那么如果B=11，我们所要计算的就是：

[123]2 % 11 = 123123 % 11 = 123*1000 % 11 + 123 % 11.

接着化简下去，记a = 123 % 11，m = 1000 % 11.

ans = (a * m % 11 + a) % 11.

改了下，数据对上了，好像还是wa，先贴上：

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long LL;
LL B;
int k;
char s[1010];
int dfs(LL& ans, int &i, LL& c){
    int len = strlen(s);
    for(i; i < len;){
        if(k == 0)c = 1;
        if(s[i] == ']'){
            i = i + 2;
            k--;
            return s[i - 1] - '0';
        }
        else if(s[i] == '['){
            LL x = 0;
            k++;
            i++;
            int cnt = dfs(x, i, c);
        //    cout << c << " " << cnt << " " << x << endl;
            LL temp = 1;
            while(cnt--){
                ans = ans * c % B + x % B;
                temp = temp * c % B;
                ans %= B;
            }
            c = temp;
        }
        else ans = (ans * 10 + s[i] - '0')%B, c = (c * 10)%B, i++;    
    }
}
int main(){
    int T;
    cin >> T;
    while(T--){
        scanf("%s%lld", s, &B);
        int i = 0;
        LL ans = 0;
        LL c = 1;
        k = 0;
        dfs(ans, i, c);
        printf("%lld\n", ans);
    }
    return 0;
}

超内存：

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
typedef long long LL;
int dfs(string& ans, string s, int &cur){
    for(int i = cur; i < s.length();){
        if(s[i] == ']'){
            cur = i + 2;
            return s[i + 1] - '0';
        }
        else if(s[i] == '['){
            string x;
            i++;
            int cnt = dfs(x, s, i);
            //cout << cnt << endl;
            while(cnt--){
                ans += x;
            }
        }
        else ans += s[i], i++;    
    }
}
int main(){
    int T;
    string s;
    LL B;
    cin >> T;
    while(T--){
        cin >> s;
        cin >> B;
        int i = 0;
        string A;
        dfs(A, s, i);
        LL temp = 0;
        for(int i = 0; i < A.length(); i++){
            temp = temp * 10 + A[i] - '0';
            if(temp > B){
                temp = temp % B;
            }
        }
        printf("%lld\n", temp);
    }
    return 0;
}