HDU 4608 I-number（模拟）

I-number

Time Limit: 5000ms   Memory limit: 65536K  有疑问？点这里^_^

题目描写叙述

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.

输入

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.

输出

Output the I-number of x for each query.

演示样例输入

1
202

演示样例输出

208

提示

来源

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;
}