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.
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.
3 [0]9[[1]2]3 10007 [[213231414343214231]5]1 10007 [0012]1 1Sample Output
1034 3943
0
题意 : 给定一个字符串,它表示一个数,表示方法是 [12]x , x是个位数,表示重复次数,给b求该数取模b
#include <cstdio> #include <cmath> #include <iostream> #include <cstring> #include <string> using namespace std; typedef long long ll; char s[1000006]; ll md; ll qm(ll a,ll n) { ll ans = 1; while(n) { if(n%2) { ans = (ans*a) % md; } a = (a*a)%md; n /= 2; } return ans; } void dfs(int l,int r,ll &k,ll &v) { k = 0; v = 0; int b = 0; int x,y; for(int i = l; i <= r; i++) { if(s[i] == '[') { if(b == 0) x = i+1; b++; } else if(s[i] == ']') { b--; if(b == 0) y = i-1; else continue; i++; ll t = s[i]-'0'; ll ck,cv; dfs(x,y,ck,cv); k += ck*t; ll e = qm(10,ck); while(t--) v = (v*e+cv)%md; } else if(!b) { v = (v*10+s[i]-'0')%md; k++; } } } int main() { int t; scanf("%d",&t); while(t--) { scanf("%s%d",s,&md); ll k,v; int len = strlen(s); dfs(0,len-1,k,v); cout << v <<endl; } return 0; }