I-number

 

I-number

Time Limit: 5000MS Memory limit: 65536K

题目描述

 

 题目描述
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.
示例输入
1
202

示例输出
208

提示

题目意思：

输入x,输出y;

1.y>x;

2.y各位和是10的整倍数

3注意x，这是一个大数问题

#include<cstdio>
#include<cstring>
int s[100005];
int a(int n)//把数字+1，同时返回+1后的数字的各位数字和
{
    int i,sum;
    s[n-1]++;
    i=n-1;
    while(s[i]>9&&i>0)//判断>9就连续进位，最后一位不进位
    {
        s[i]-=10;
        s[--i]+=1;
    }
    sum=0;
    i=n-1;
    while(i>0)
        sum+=s[i--];
    if(s[0]<10)//最前面一位注意<10,就直接加进去
        sum+=s[0];
    else//否则就模拟进位
    {
        int a=s[0];
        while(a)
        {
            sum+=a%10;
            a/=10;
        }
    }
    return sum;
}
int main()
{
    char p[100005];
    int n,len,i,sum;
    scanf("%d",&n);
    while(n--)
    {
        scanf("%s",p);
        len=strlen(p);
        for(i=0; i<len; i++)
            s[i]=(p[i]-'0');//字符转数字
        sum=a(len);//数字+1；
        while(sum%10!=0)
            sum=a(len);//持续+1；
        for(i=0; i<len; i++)
            printf("%d",s[i]);
        printf("\n");
    }
    return 0;
}