hdu 4956 Poor Hanamichi(枚举)

acm_cxq

于 2016-07-01 19:06:49 发布

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分类专栏：枚举文章标签： ACM HDU

本文链接：https://blog.youkuaiyun.com/acm_cxq/article/details/51803809

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本文介绍了一道关于求解特定范围内数字属性的问题，通过分析数字的奇偶位数之和来找出符合特定条件的数字数量，并提供了一种验证解决方案正确性的方法。

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Poor Hanamichi

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 895 Accepted Submission(s): 408

Problem Description

Hanamichi is taking part in a programming contest, and he is assigned to solve a special problem as follow: Given a range [l, r] (including l and r), find out how many numbers in this range have the property: the sum of its odd digits is smaller than the sum of its even digits and the difference is 3.

A integer X can be represented in decimal as:

X=An×10n+An−1×10n−1+…+A2×102+A1×101+A0
The odd dights are

A1,A3,A5… and

A0,A2,A4… are even digits.

Hanamichi comes up with a solution, He notices that:

102k+1 mod 11 = -1 (or 10),

102k mod 11 = 1,
So X mod 11
=

(An×10n+An−1×10n−1+…+A2×102+A1×101+A0)mod11
=

An×(−1)n+An−1×(−1)n−1+…+A2−A1+A0
= sum_of_even_digits – sum_of_odd_digits
So he claimed that the answer is the number of numbers X in the range which satisfy the function: X mod 11 = 3. He calculate the answer in this way :
Answer = (r + 8) / 11 – (l – 1 + 8) / 11.

Rukaw heard of Hanamichi’s solution from you and he proved there is something wrong with Hanamichi’s solution. So he decided to change the test data so that Hanamichi’s solution can not pass any single test. And he asks you to do that for him.

Input

You are given a integer T (1 ≤ T ≤ 100), which tells how many single tests the final test data has. And for the following T lines, each line contains two integers l and r, which are the original test data. (1 ≤ l ≤ r ≤

1018 )

Output

You are only allowed to change the value of r to a integer R which is not greater than the original r (and R ≥ l should be satisfied) and make Hanamichi’s solution fails this test data. If you can do that, output a single number each line, which is the smallest R you find. If not, just output -1 instead.

Sample Input

Sample Output

思路：直接枚举即可

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
int check(long long k)
{
    long long l=k;
    int num=0;
    while(l)
    {
        l/=10;
        num++;
    }
    int lsum=0,rsum=0,flag=0;
    while(k)
    {
        if(!flag) rsum+=k%10;
        else lsum+=k%10;
        k/=10;
        flag=!flag;
    }
    if(rsum==lsum+3) return 1;
    return 0;
}
int main()
{
    int T;
    long long l,r;
    scanf("%d",&T);
    while(T--)
    {
            scanf("%lld %lld",&l,&r);
            long long num=0,v;
            long long flag=0;
            for(long long i=l; i<=r; i++)
            {
                if(check(i)) num++;
                v=(i+8)/11-(l+7)/11;
                if(v!=num)
                {
                    flag=i;
                    break;
                }
            }
            if(flag) printf("%lld\n",flag);
            else printf("-1\n");
    }
    return 0;
}