hdu 4068 I-number【大数】

题目：

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

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=27048#problem/G

2013暑期多校联合训练——80+高校，300+队伍，10000元奖金，敬请期待~

I-number

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

Total Submission(s): 1006    Accepted Submission(s): 398

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

 

Source

2013 Multi-University Training Contest 1

 

Recommend

liuyiding

 

题意：

第一个数 T 代表测试数据组数

每组给你一个大数 N （N的长度 <= 100000）

求最小的 > N 且满足每一位相加的总和能够整除 10 的数

算法：

大数相加,只是+1比较简单，随便模拟一下就好了

思路：

不断的 + 1 直到满足情况

官方题解中也说的是最多+20个 1 就可以求出

昨天白天比赛的时候是 浩神 AC 的，KB 神也和我说了下怎么做, 白天看了下 浩神的代码，自己写，还是要 WA

昨晚比赛时按照浩神的思路纠结出了 AC的代码

刚刚问了下 KB 神，原来是掉了个初始化，改了也 AC 了

code：

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;

const int maxn = 200000;
int a[maxn];
char str[maxn];
int len;

void add()
{
    int c = 1;
    for(int i = 0;; i++)
    {
        int tmp = a[i]+c;
        a[i] = tmp%10;
        c = tmp/10;
        if(c == 0) break; //加到没有进位
    }
    if(a[len] != 0) len++; //加到头有进位
}

bool judge()
{
    int sum = 0;
    for(int i = 0; i < len; i++)
        sum += a[i];
    return sum%10;
}

int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%s", str);
        len = strlen(str);

        memset(a,0,sizeof(a)); //不能少否则会WA，前面的组a[len]可能会有价
        int j = len;
        for(int i = 0; i < len; i++) a[i] = str[--j]-'0';
        add();
        while(judge()) add();

        for(int i = len-1; i >= 0; i--) printf("%d",a[i]);
        printf("\n");
    }
    return 0;
}

#include<stdio.h>
#include<string.h>

const int maxn = 200000;
int a[maxn];
char str[maxn];
int len;

void add()
{
    int c = 1;
    for(int i = 0; i < len; i++)
    {
        int tmp = a[i]+c;
        a[i] = tmp%10;
        c = tmp/10;
        if(c == 0) return;
    }
    a[len] = c; len++;
    return;
}

int judge()
{
    int sum = 0;
    for(int i = 0; i < len; i++)
    {
        sum += a[i];
    }
    return sum%10;
}
int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%s", str);
        len = strlen(str);

        int j = len;
        for(int i = 0; i < len; i++) a[i] = str[--j]-'0';
        add();
        while(judge()) add();

        for(int i = len-1; i >= 0; i--) printf("%d", a[i]);
        printf("\n");
    }
}