The I-number of x is defined to be an integer y, which satisfied the the conditions below:

1.  y>x;

2.  the sum of each digit of y(under base 10) is the multiple of 10;

3.  among all integers that satisfy the two conditions above, y shouble be the minimum.

Given x, you\'re required to calculate the I-number of x.

An integer T(T≤100) will exist in the first line of input, indicating the number of test cases.

The following T lines describe all the queries, each with a positive integer x. The length of x will not exceed 10⁵.

Output the I-number of x for each query.

1

202

208

2013 Multi-University Training Contest 1

 

我不得不说这道题真的非常坑，我要是不搜题解预计是毁掉了 本来题意非常easy，就是给定一个x（因为可能非常大用字符串模拟），求出比x大的且各位数字之和为10的倍数的数，

要求输出符合上诉条件的最小数。

but,这个x居然是能够有前导0的（它没说）也就是说输入 00202 输出 00208

 

代码略挫QAQ

#include <stdio.h>

#include <iostream>

#include <algorithm>

#include <string.h>

#include <cstdio>

#include <cctype>

using namespace std;

char x[100010];

int digitsum()

{

	int sum=0;

	for(int i=0;i<strlen(x);i++)

		sum+=(x[i]-'0');

	return sum;

}

int is_o()

{

	for(int i=0;i<strlen(x);i++)

		if(x[i]!='0') return 0;

	return 1;

}

int main()

{

 int i,T;

 cin>>T;

 getchar();

 while(T--)

 {

 	cin>>x;

 	int s=-99;

 	while(s%10)

	{

		int p=strlen(x)-1;

		int len=strlen(x);

		x[p]++;

		while(x[p]>'9')

		{

			x[p--]-=10;

			if(p>=0)

			x[p]++;

		}

		if(is_o())

		{

			x[0]='1';

			for(i=1;i<=len;i++)

				x[i]='0';

			x[i]='\0';

		}

		s=digitsum();

	}

	cout<<x<<endl;

 }

  return 0;

}