Ability To Convert

題意：給你一個n進制的60位的數，但是由於Alexander只會寫0->9,所以他就會用10來表示十而不是A(假設進制>10);

題解：模擬就好了，先走往前走進制的位數的倍數，再判斷一下首位是不是0，或者這個數有沒有大於等於進制，如果有就不行，就要往後走，走到一個非0開頭的點，或者就是走到只有1個0的點，然後算出目前這段區間的答案加一個倍數就好了。

代碼：

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
 4
 
 #define LL long long
 5 #define ULL unsigned LL
 6 #define fi first
 7 #define se second
 8 #define pb push_back
 9 #define lson l,m,rt<<1
10 #define rson m+1,r,rt<<1|1
11 #define max3(a,b,c) max(a,max(b,c))
12 #define min3(a,b,c) min(a,min(b,c))
13 typedef pair<int,int> pll;
14 const
 
 int INF = 0x3f3f3f3f;
15 const LL mod =  (int)1e9+7;
16 const int N = 2e5 + 100;
17 char str[N];
18 char cmp[N];
19 LL ans = 0;
20 LL tmp = 0;
21 LL n = 0;
22 LL base = 1;
23 LL b = 1;
24 LL t;
25 vector<LL> vc;
26 int main(){
27     scanf("%s", cmp);
28     int len1 = strlen(cmp);
29     for(int i = 0
 
; i < len1; i++){
30         n = n*10 + cmp[i] - ‘0‘;
31     }
32     scanf("%s", str);
33     if(str[0] == ‘0‘){
34         cout << 0 << endl;
35         return 0;
36     }
37     int len2 = strlen(str);
38     for(int i = len2-1; i >= 0;){
39         int j = i - len1 + 1;
40         if(j < 0) j = 0;
41         //cout << j <<‘ ‘<< i << endl;
42         if((j+len1-1 == i && strcmp(cmp,str+j) <= 0) || str[j] == ‘0‘){
43             j++;
44             if(j >= i) j = i;
45             while(str[j] == ‘0‘ && j < i) j++;
46             t = 0;
47             for(int k = j; k <= i; k++)
48                 t = t*10 + str[k] - ‘0‘;
49             ans += b*t;
50             b *= n;
51         }
52         else {
53             t = 0;
54             for(int k = j; k <= i; k++)
55                 t = t*10 + str[k] - ‘0‘;
56             ans += b*t;
57             b *= n;
58         }
59         i = j - 1;
60     }
61     printf("%I64d", ans);
62     return 0;
63 }

CodeForces 758 D Ability To Convert