2017 ACM-ICPC 亚洲区（西安赛区）网络赛 C. Sum

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

Input Format

First line an integer TTT,
indicates the number of test cases (T≤100T \le 100T≤100).
Then Each line has a single integer x(1≤x≤1000000)x(1 \le x \le 1000000)x(1≤x≤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
200020002000.
If there are more than one answer, output anyone is ok.

样例输入

1
1

样例输出

89999999999999999999999999

题目来源

2017 ACM-ICPC 亚洲区（西安赛区）网络赛

题意：给你一个x，要你求一个正整数k，使得k*x的结果的每一位的和为233的倍数。

解题思路：这题有bug，直接输出0也是对的……但我们不这么做，实际上，k可以只由1组成，然后暴力枚举所有由1组成的数，直到出答案为止。这里要用到大数模板，我就上网百度了一份。感谢某位大神的大数模板……只看main函数就可以了

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <algorithm>
using namespace std;

const int MAXN = 2000;

struct bign
{
int len, s[MAXN];
bign ()
{
memset(s, 0, sizeof(s));
len = 1;
}
bign (int num) { *this = num; }
bign (const char *num) { *this = num; }
bign operator = (const int num)
{
char s[MAXN];
sprintf(s, "%d", num);
*this = s;
return *this;
}
bign operator = (const char *num)
{
for(int i = 0; num[i] == '0'; num++) ; //去前导0
len = strlen(num);
for(int i = 0; i < len; i++) s[i] = num[len-i-1] - '0';
return *this;
}
bign operator + (const bign &b) const //+
{
bign c;
c.len = 0;
for(int i = 0, g = 0; g || i < max(len, b.len); i++)
{
int x = g;
if(i < len) x += s[i];
if(i < b.len) x += b.s[i];
c.s[c.len++] = x % 10;
g = x / 10;
}
return c;
}
bign operator += (const bign &b)
{
*this = *this + b;
return *this;
}
void clean()
{
while(len > 1 && !s[len-1]) len--;
}
bign operator * (const bign &b) //*
{
bign c;
c.len = len + b.len;
for(int i = 0; i < len; i++)
{
for(int j = 0; j < b.len; j++)
{
c.s[i+j] += s[i] * b.s[j];
}
}
for(int i = 0; i < c.len; i++)
{
c.s[i+1] += c.s[i]/10;
c.s[i] %= 10;
}
c.clean();
return c;
}
bign operator *= (const bign &b)
{
*this = *this * b;
return *this;
}
bign operator - (const bign &b)
{
bign c;
c.len = 0;
for(int i = 0, g = 0; i < len; i++)
{
int x = s[i] - g;
if(i < b.len) x -= b.s[i];
if(x >= 0) g = 0;
else
{
g = 1;
x += 10;
}
c.s[c.len++] = x;
}
c.clean();
return c;
}
bign operator -= (const bign &b)
{
*this = *this - b;
return *this;
}
bign operator / (const bign &b)
{
bign c, f = 0;
for(int i = len-1; i >= 0; i--)
{
f = f*10;
f.s[0] = s[i];
while(f >= b)
{
f -= b;
c.s[i]++;
}
}
c.len = len;
c.clean();
return c;
}
bign operator /= (const bign &b)
{
*this = *this / b;
return *this;
}
bign operator % (const bign &b)
{
bign r = *this / b;
r = *this - r*b;
return r;
}
bign operator %= (const bign &b)
{
*this = *this % b;
return *this;
}
bool operator < (const bign &b)
{
if(len != b.len) return len < b.len;
for(int i = len-1; i >= 0; i--)
{
if(s[i] != b.s[i]) return s[i] < b.s[i];
}
return false;
}
bool operator > (const bign &b)
{
if(len != b.len) return len > b.len;
for(int i = len-1; i >= 0; i--)
{
if(s[i] != b.s[i]) return s[i] > b.s[i];
}
return false;
}
bool operator == (const bign &b)
{
return !(*this > b) && !(*this < b);
}
bool operator != (const bign &b)
{
return !(*this == b);
}
bool operator <= (const bign &b)
{
return *this < b || *this == b;
}
bool operator >= (const bign &b)
{
return *this > b || *this == b;
}
string str() const
{
string res = "";
for(int i = 0; i < len; i++) res = char(s[i]+'0') + res;
return res;
}
};

istream& operator >> (istream &in, bign &x)
{
string s;
in >> s;
x = s.c_str();
return in;
}

ostream& operator << (ostream &out, const bign &x)
{
out << x.str();
return out;
}

int main()
{
bign x, k;

int t;
scanf("%d",&t);
while(t--)
{
cin>>x;

k="1";
while(1){

bign temp=x*k;

int sum=0;
for(int i=0;i<temp.len;i++)
sum+=temp.s[i];

if(sum%233==0){
cout<<k.str()<<endl;
break;
}

k*=10;
k+=1;

}

}
return 0;
}