hdoj4956Poor Hanamichi【暴力】
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Poor Hanamichi
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 864 Accepted Submission(s): 383
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 areA1,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.
A integer X can be represented in decimal as:
The odd dights are
Hanamichi comes up with a solution, He notices that:
So X mod 11
=
=
= 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
33 42 507 83
Sample Output
-1-180
Source
BestCoder Round #5
直接暴力即可
#include<cstdio>#include<cstdlib>#include<cstring>#include<algorithm>#include<cmath>#include<list>#include<queue>#include<vector>using namespace std;bool judge(long long n){long long odd=0,even=0;int flag=1;while(n){if(flag==1)even+=n%10;else odd+=n%10;n/=10;flag*=-1;}return even-odd==3;}int main(){int t;long long l,r,i;scanf("%d",&t);while(t--){scanf("%lld%lld",&l,&r);long long cnt=0;for(i=l;i<=r;++i){long long num=(i+8)/11-(l-1+8)/11;if(judge(i))cnt++;if(cnt!=num)break;}if(i<=r){printf("%lld\n",i);}else {printf("-1\n");}}return 0;}
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