HDU 4608 I-number (数学&字符串处理)

I-number

http://acm.hdu.edu.cn/showproblem.php?pid=4608

Time Limit: 10000/5000 MS (Java/Others)

Memory Limit: 32768/32768 K (Java/Others)

Problem Description

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.

Input

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 10 ⁵.

Output

Output the I-number of x for each query.

Sample Input

1 202

Sample Output

208

题意：求一个大于x的且最接近x的数y，y所有数位上的数字和sum是10的倍数。

思路：由于x的长度可达10^5，采用字符串处理即可。

关键是判断能不能在不进位（只改变最后一个数）的时候找到y，此时把最后一位加上10 - sum%10就行。（需满足条件：最后一位

若最后一位进位，就先判断仅为之后所有位之和，比如57999进位后变58000，sum%10=3，那么最后一位加上10 - sum%10就行

特判：所有位数为9，那直接先输出一个1，然后把x最后一位变成9，剩下变成0就行。

完整代码：

/*171ms,332KB*/

#include
#include

char x[100002];

int main()
{
	int t;
	scanf("%d", &t);
	getchar();
	while (t--)
	{
		memset(x, 0, sizeof(x));
		gets(x);
		int len = strlen(x), sum = 0, i;
		for (int i = 0; i < len; ++i)
			sum += x[i] - '0';
		sum %= 10;
		if (x[len - 1] - '0' < sum)
			x[len - 1] += 10 - sum;
		else
		{
			x[len - 1] = '9';
			sum = 0;
			for (i = len - 1; i >= 0; --i)
				if (x[i] != '9')
					break;
				else
					x[i] = '0';
			if (i == -1)
			{
				x[len - 1] = '9';
				putchar('1');
				printf("%s\n", x);
				continue;
			}
			++x[i];///进位
			for (int j = 0; j <= i; ++j)
				sum += x[j] - '0';
			sum %= 10;
			if (sum)
				x[len - 1] += 10 - sum;
		}
		printf("%s\n", x);
	}
	return 0;
}