HDU 4608 I-number

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

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I-number

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 3092    Accepted Submission(s): 1163

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 10⁵.

 

Output

Output the I-number of x for each query.

 

Sample Input

1
202

 

Sample Output

208

 

题目大意：给你一个大整数x,叫你找出一个比x大的最小的数y，它满足条件：y的各个数位之和能被10整除

大数水题，不断模拟加1即可，不用想得太复杂。o(n)的复杂度喽

#include<cstdio>#include<cstring>#define maxn 100005char a[maxn],b[maxn];int na[maxn],nb[maxn];void add(){    memset(na,0,sizeof(na));    memset(nb,0,sizeof(nb));    int la=strlen(a),lb=strlen(b);    for(int i=0;a[i];i++) na[la-1-i]=a[i]-'0';    for(int i=0;b[i];i++) nb[lb-1-i]=b[i]-'0';    int lmax=la>lb?la:lb;    for(int i=0;i<lmax;i++) na[i]+=nb[i],na[i+1]+=na[i]/10,na[i]%=10;    if(na[lmax]) lmax++;    for(int i=0;i<lmax;i++) a[lmax-i-1]=na[i]+'0';    a[lmax]='\0';}bool judge(){    int n=0;    for(int i=0;a[i];i++) n+=a[i]-'0';   // printf("n=%d\n",n);    return n%10;}int main(){    int t;    scanf("%d",&t);    while(t--)    {        scanf("%s",a);        strcpy(b,"1");add();        while(judge()) strcpy(b,"1"),add();        printf("%s\n",a);    }    return 0;}