hdu 4956(思路题)

Poor Hanamichi

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 893    Accepted Submission(s): 407

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

3 3 4 2 50 7 83

 

 

Sample Output

-1 -1 80

 

题意:现在想得到[l,r]区间里面有多少个数字的偶数位之和比奇数位大 3，然后有人就想了个办法，然后bilibilibilibili得到一个公式来计算ans = (r+8)/11-(l+1-8)/11

现在有人怀疑他这样做是错的，然后想来验证这种做法,于是给出区间[l,r]，验证这段区间是否上述公式，如果满足，输出-1，不满足，输出不满足的最小的那个数字。

题解:对 [l,r]里面的每个数进行判断,先用公式求一遍，然后再判断一下这个数，如果公式所得不等于当前暴力枚举的结果，直接break

#include <stdio.h>
#include <math.h>
#include <iostream>
#include <algorithm>
#include <string.h>
#include <vector>
using namespace std;
typedef long long LL;

bool judge(LL x){
    LL bit[20];
    int t = 0;
    while(x){
        bit[t++] = x%10;
        x/=10;
    }
    LL odd=0,even=0;
    for(int i=0;i<t;i+=2){
        odd+=bit[i];
    }
    for(int i=1;i<t;i+=2){
        even+=bit[i];
    }
    if(odd-even==3) return true;
    return false;
}
int main()
{
    int tcase;
    scanf("%d",&tcase);
    while(tcase--){
        LL l,r;
        scanf("%lld%lld",&l,&r);
        LL ans = -1,x=0;
        for(LL i=l;i<=r;i++){
            LL temp = (i+8)/11- (l-1+8)/11;
            if(judge(i)){
                x++;
            }
            if(x!=temp){
                ans = i;
                break;
            }
        }
        printf("%lld\n",ans);
    }
}

 

转载于:https://www.cnblogs.com/liyinggang/p/5629299.html