2017 ACM-ICPC 亚洲区（西安赛区）网络赛 C.Sum

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Define the function $S(x)$ for $x$ is a positive integer. $S(x)$ equals to the sum of all digit of the decimal expression of $x$. Please find a positive integer $k$ that $S(k\ast x)\%233=0$.

Input Format

First line an integer $T$, indicates the number of test cases ($T\le 100$). Then Each line has a single integer $x(1\le x\le 1000000)$ indicates i-th test case.

Output Format

For each test case, print an integer in a single line indicates the answer. The length of the answer should not exceed $2000$. If there are more than one answer, output anyone is ok.

样例输入

11

样例输出

89999999999999999999999999

题目来源

2017 ACM-ICPC 亚洲区（西安赛区）网络赛

搞笑，五个小时做不出来的题，听说每次都输出的是0就过了，0是正整数吗？搞笑，赛后提交全输出0就不对。。。。。仔细想想输出233个9就行，这样的数乘任何数的和都是233的倍数，别问我是怎么知道的，我不知道，这个题我没看过，这个题什么意思？西安网络赛是什么鬼？为什么入坑ACM？洗洗睡吧，明天还有青岛站。

代码实现：

#include<iostream>#include<algorithm>#include<cstring>#include<cmath>#include<queue>#include<cstdio>#define ll long long#define mset(a,x) memset(a,x,sizeof(a))using namespace std;const double PI=acos(-1);const int inf=0x3f3f3f3f;const double esp=1e-6;const int maxn=100005;const int mod=1e9+7;int dir[4][2]={0,1,1,0,0,-1,-1,0};ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}ll lcm(ll a,ll b){return a/gcd(a,b)*b;}ll inv(ll b){if(b==1)return 1; return (mod-mod/b)*inv(mod%b)%mod;}ll fpow(ll n,ll k){ll r=1;for(;k;k>>=1){if(k&1)r=r*n%mod;n=n*n%mod;}return r;}int main(){int t,n,i,j;cin>>t;while(t--){cin>>n;for(i=0;i<233;i++)cout<<"9";cout<<endl;}return 0;}