AtCoder Beginner Contest 363(A~D题)

A - Piling Up

 思路:

我们只需要找到下一阶段的下限。a / 100 是本阶段+1 变成下一阶段，再 * 100变成下限，再与原来的相减即可。

代码:

#include<bits/stdc++.h>
using namespace std;
#define N 200010
typedef long long ll;
typedef unsigned long long ull;
ll n, m, h, k, t, x, y, z;
ll a, b, c, d, mod = 998244353;
ll ans, num, sum1=0, sum, sum2=0, cnt, maxx, minn = 1e9;
ll f1[N], f2[N], dp[N], an[N];
bool flag, vis[N];
string s,s1,s2;
map<ll, ll>mp;
int main()
{
    cin >> n;
    if (n % 100 == 0)
    {
        cout << "100\n";
        return 0;
    }
    else
    {
        k = n / 100;
        k = (k + 1) * 100;
        cout << k - n << "\n";
    }
	return 0;
}

B - Japanese Cursed Doll

思路:

将L