ZOJ Problem Set -- 4085 Little Sub and Mr.Potato's Math Problem

1. 题目来源：ZOJ Problem Set -- 4085  Little Sub and Mr.Potato's Math Problem

题意：Q(n，k)=m

将n个数按照字典序排序，k所在的位置为m

现在给出k，m 求最小的n

思路：

当k为10的整数倍，那么它一定在第 log以10为底k 的位置上

随后统计当n==k的时候，排在k前面的个数，和m比较，判断是否合法

接着不断增加n，统计每次增长排在k前面的个数，随后输出n

自己的代码;

#include 
#include 
#include 
#include 
#include 
#include 
#include 

大佬的代码：

#include

using namespace std;

typedef long long ll;
typedef unsigned long long ull;

const double eps = 1e-8;
const ll MOD = 1e9 + 7;
const ll INFLL = 0x3f3f3f3f3f3f3f3f;
const int INF = 0x3f3f3f3f;
const int maxn = 1e5 + 10;

ll k, m;
int arr[maxn];
ll pow_10[10];

void Init()
{
    pow_10[0] = 1;
    for (int i = 1; i <= 18; ++i)
    {
        pow_10[i] = pow_10[i - 1] * 10;
    }
}

void solve()
{
    ll num = 1;
    for (int i = 1;; ++i)
    {
        if (num > k) break;
        else if (num == k)
        {
            if (i == m)
            {
                printf("%lld\n", k);
                return;
            }
            else
            {
                puts("0");
                return;
            }
        }
        num *= 10;
    }
    int len = 0;
    num = k;
    while (num)
    {
        arr[++len] = num % 10;
        num /= 10;
    }
    reverse(arr + 1, arr + 1 + len);
    ll ans = 0;
    num = 0;
    for (int i = 1; i <= len; ++i)
    {
        num = num * 10 + arr[i];
        ans += num - pow_10[i - 1];
        if (i != len) ++ans;
    }
    if (ans >= m)
    {
        puts("0");
        return;
    }
    else if (ans == m - 1)
    {
        printf("%lld\n", k);
        return;
    }
    while (1)
    {
        len++;
        num *= 10;
        if (ans + num - pow_10[len - 1] >= m - 1)
        {
            ans = pow_10[len - 1] + m - ans - 2;
            printf("%lld\n", ans);
            return;
        }
        ans += num - pow_10[len - 1];
    }
}

void RUN()
{
    Init();
    int t;
    scanf("%d", &t);
    while (t--)
    {
        scanf("%lld %lld", &k, &m);
        solve();
    }
}

int main()
{
#ifdef LOCAL_JUDGE
    freopen("Text.txt", "r", stdin);
#endif // LOCAL_JUDGE

    RUN();

#ifdef LOCAL_JUDGE
    fclose(stdin);
#endif // LOCAL_JUDGE
    return 0;
}