已经菜到不行了 PAT 1010. Radix (25)

https://www.patest.cn/contests/pat-a-practise/1010

题目大意：

输入四个数字，a,b,c,d。

a和b是两个数字，c=1表示是第一个数字，c=2表示是第二个数字，d表示该数字是几进制

问：a和b是否能在某一进制下相等

思路：

傻逼二分...

一年没写过了，细节思考竟然差了这么多，尴尬了...

 //看看会不会爆int!数组会不会少了一维！
 //取物问题一定要小心先手胜利的条件
 #include <bits/stdc++.h>
 using namespace std;
 #pragma comment(linker,"/STACK:102400000,102400000")
 #define LL long long
 #define ALL(a) a.begin(), a.end()
 #define pb push_back
 #define mk make_pair
 #define fi first
 #define se second
 #define haha printf("haha\n")
 const int maxn = 1e5 + ;
 LL val;
 char ch1[maxn], ch2[maxn];

 LL get_int(char ch){
     if (ch >= '' &&  ch <= '') return 1LL * (ch - '');
     return 1LL * (ch - 'a' + );
 }
 //c 1100 1 26
 bool flag;
 LL cal(LL x, int n){
     LL ans = ;

     while (n){
         if (n & ) ans *= x;
         x = x * x;
         if (x <  || ans < ) flag = false;
         n /= ;
     }
     return ans;
 }

 LL get_val(vector<LL> ve, int radix){
     LL sum = ;
     int cnt = ;
     for (int i = ve.size() - ; i >= ; i--){
         sum = sum + ve[i] * cal(1LL * radix, cnt);
         cnt++;
     }
     if (sum < ) flag = false;
     return sum;
 }

 bool solve(vector<LL> b){
     LL lb = , rb = val + ;
     for (int i = ; i < b.size(); i++)
         lb = max(lb, b[i] + );
     while (lb < rb){
         LL mid = (lb + rb) / ;
         flag = true;
         LL tmp = get_val(b, mid);
         if (!flag) {
             rb = mid - ; continue;
         }
         if (val > tmp) lb = mid + ;
         if (val < tmp) rb = mid - ;
         if (val == tmp) rb = mid;
     }
     if (get_val(b, lb) == val) {
         printf("%lld\n", lb);
         return true;
     }
     return false;
 }

 int main(){
     scanf("%s%s", ch1, ch2);
     int ty, change;
     scanf("%d%d", &ty, &change);
     vector<LL> a, b;
     if (ty == ){
         for (int i = ; ch1[i] != '\0'; i++)
             a.push_back(get_int(ch1[i]));
         for (int i = ; ch2[i] != '\0'; i++)
             b.push_back(get_int(ch2[i]));
     }
     else {
         for (int i = ; ch2[i] != '\0'; i++)
             a.push_back(get_int(ch2[i]));
         for (int i = ; ch1[i] != '\0'; i++)
             b.push_back(get_int(ch1[i]));
     }
     val = get_val(a, change);
     if(!solve(b)) puts("Impossible");
     return ;
 }