CodeForces 808A——Lucky Year——思维，模拟

A. Lucky Year

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Apart from having lots of holidays throughout the year, residents of Berland also have whole lucky years. Year is considered lucky if it has no more than 1 non-zero digit in its number. So years 100, 40000, 5 are lucky and 12, 3001 and 12345 are not.

You are given current year in Berland. Your task is to find how long will residents of Berland wait till the next lucky year.

Input

The first line contains integer number n (1 ≤ n ≤ 109) — current year in Berland.

Output

Output amount of years from the current year to the next lucky one.

Examples

input

4

output

1

input

201

output

99

input

4000

output

1000

Note

In the first example next lucky year is 5. In the second one — 300. In the third — 5000.

直接求出大于等于n的幸运年再减就好了，pow返回值是double型的还是不太敢用。

#include<cstdio>#include<iostream>#include<algorithm>#include<cmath>#include<cstring>#include<vector>#include<set>#include<queue>#include<map>#include<stack>#define max3(x, y, z) max((x), max((y), (z)))#define min3(x, y, z) min((x), min((y), (z)))#define pb push_back#define mp make_pairusing namespace std;const int N = 1000;int cnt;    ///获得输入数字的位数/// 获取输入数字最高位的数字int g(int x){    int t;    while(x){        t = x % 10;        cnt ++;        x /= 10;    }    return t;}/// 10的x次方int p(int x){    int ans = 1;    for(int i = 0; i < x; i ++){        ans *= 10;    }    return ans;}int main(){    int n;    int d;    while(cin >> n){        cnt = 0;        d = g(n);        //cout << (d + 1) * p(cnt - 1) - n << endl;        printf("%.lf\n", (d + 1) * pow(10, (cnt - 1)) - n);    }    return 0;}

偶然看到别人妖艳不做作的打表

#include <bits/stdc++.h>using namespace std;typedef long long ll;int main(){    ll n;    while(scanf("%lld",&n)!=EOF)    {        if(n>=0&&n<=9)        cout<<1<<endl;        else if(n>=10&&n<=99)            cout<<(n/10+1)*10-n<<endl;        else if(n>=100&&n<=999)            cout<<(n/100+1)*100-n<<endl;        else if(n>=1000&&n<=9999)            cout<<(n/1000+1)*1000-n<<endl;        else if(n>=10000&&n<=99999)            cout<<(n/10000+1)*10000-n<<endl;        else if(n>=100000&&n<=999999)            cout<<(n/100000+1)*100000-n<<endl;        else if(n>=1000000&&n<=9999999)            cout<<(n/1000000+1)*1000000-n<<endl;        else if(n>=10000000&&n<=99999999)            cout<<(n/10000000+1)*10000000-n<<endl;        else if(n>=100000000&&n<=999999999)            cout<<(n/100000000+1)*100000000-n<<endl;        else if(n>=1000000000&&n<=9999999999)            cout<<(n/1000000000+1)*1000000000-n<<endl;    }    return 0;}