CodeForces 758 D Ability To Convert

Ability To Convert

题意：给你一个n进制的60位的数，但是由于Alexander只会写0->9,所以他就会用10来表示十而不是A(假设进制>10);

题解：模拟就好了，先走往前走进制的位数的倍数，再判断一下首位是不是0，或者这个数有没有大于等于进制，如果有就不行，就要往后走，走到一个非0开头的点，或者就是走到只有1个0的点，然后算出目前这段区间的答案加一个倍数就好了。

代码：

 #include<bits/stdc++.h>
 using namespace std;
 #define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
 #define LL long long
 #define ULL unsigned LL
 #define fi first
 #define se second
 #define pb push_back
 #define lson l,m,rt<<1
 #define rson m+1,r,rt<<1|1
 #define max3(a,b,c) max(a,max(b,c))
 #define min3(a,b,c) min(a,min(b,c))
 typedef pair<int,int> pll;
 const int INF = 0x3f3f3f3f;
 const LL mod =  (int)1e9+;
 const int N = 2e5 + ;
 char str[N];
 char cmp[N];
 LL ans = ;
 LL tmp = ;
 LL n = ;
 LL base = ;
 LL b = ;
 LL t;
 vector<LL> vc;
 int main(){
     scanf("%s", cmp);
     int len1 = strlen(cmp);
     for(int i = ; i < len1; i++){
         n = n* + cmp[i] - '';
     }
     scanf("%s", str);
     if(str[] == ''){
         cout <<  << endl;
         return ;
     }
     int len2 = strlen(str);
     for(int i = len2-; i >= ;){
         int j = i - len1 + ;
         if(j < ) j = ;
         //cout << j <<' '<< i << endl;
         if((j+len1- == i && strcmp(cmp,str+j) <= ) || str[j] == ''){
             j++;
             if(j >= i) j = i;
             while(str[j] == '' && j < i) j++;
             t = ;
             for(int k = j; k <= i; k++)
                 t = t* + str[k] - '';
             ans += b*t;
             b *= n;
         }
         else {
             t = ;
             for(int k = j; k <= i; k++)
                 t = t* + str[k] - '';
             ans += b*t;
             b *= n;
         }
         i = j - ;
     }
     printf("%I64d", ans);
     return ;
 }

CF 758 D