Codeforces-758D Ability To Convert

传送门

给定一个N，和一个字符串，已知字符串是由不大于N的十进制数拼接而成，从左到右分别代表N^k, N^(k-1)...., N^1, N^0的权数，求这个字符串代表的数的最小值

嗯 注意前导零

#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <queue>
#define INF 0x3f3f3f3f
using namespace std;
typedef long long LL;
LL N, M;
LL ans;
const int maxk = 66;
string s;
string ss;

void add(int i, int j) {
    LL tmp = 0;
    for (int k = i; k <= j; k++) {
        tmp = tmp * 10 + s[k] - '0';
    }
    ans = ans + tmp * M;
}

int main() {
    LL tmp = 0;
    cin >> ss;
    int l = ss.length();
    for (int i = 0 ; i <= l - 1; i++) {
        N = N * 10 + ss[i] - '0';
    }
    M = 1;
    cin >> s;
    int len = s.length();
    for (int i = len - 1; i >= 0; i -= l) {
        if (i >= l - 1) {
            int k;
            if (s.substr(i - l + 1, l) >= ss) {
                add(i - l + 2, i);
                k = i - l + 2;
            } else {
                add(i - l + 1, i);
                k = i - l + 1;
            }
            while (k <= i && s[k] == '0') {
                k++;
            }
            if (k == i + 1) {
                i = i + l - 1;
            } else {
                i = k + l - 1;
            }
        } else {
            add(0, i);
        }
        M *= N;
    }
    printf("%lld
", ans);
    return 0;
}