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2016-06-25 23:08
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暑期(二)
744:48:54
1200:00:00
Overview
Problem
Status
Rank (99849)
A B C D E F G H I J K L M N O P Q R S T
R - R
Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u
Submit Status
Description
Find the number of k-divisible numbers on the segment [a, b]. In other words you need to find the number of such integer
values x thata ≤ x ≤ b and x is divisible by k.
Input
The only line contains three space-separated integers k, a and b (1 ≤ k ≤ 1018; - 1018 ≤ a ≤ b ≤ 1018).
Output
Print the required number.
Sample Input
Input
Output
Input
Output
FAQ | About Virtual Judge | Forum | Discuss | Open
Source Project
All Copyright Reserved ©2010-2014 HUST ACM/ICPC TEAM
Anything about the OJ, please ask in the forum, or contact author:Isun, but
NEVER contact freefcw@gmail.com, who is irrelevant to this site !!
Server Time:
姐:两种情况!
1.b<0, b如果不能整除以k 。则产生物差1.
2.a>0 a如果不能整除以k。则产生误差1.
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int main()
{
long long a,b,n,i,j,l,k,t,s;
while(scanf("%lld%lld%lld",&k,&a,&b)!=EOF)
{
s=b/k-a/k+1;
if(a>=0)
{
if(a%k!=0)
s--;
}
else
if(b<0&&b%k!=0)
s--;
printf("%lld\n",s);
}
return 0;
}
PROBLEM
STATUS
CONTEST
Add
Contest
Statistic
LOGOUT
dtl666
UPDATE
暑期(二)
744:48:54
1200:00:00
Overview
Problem
Status
Rank (99849)
A B C D E F G H I J K L M N O P Q R S T
R - R
Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u
Submit Status
Description
Find the number of k-divisible numbers on the segment [a, b]. In other words you need to find the number of such integer
values x thata ≤ x ≤ b and x is divisible by k.
Input
The only line contains three space-separated integers k, a and b (1 ≤ k ≤ 1018; - 1018 ≤ a ≤ b ≤ 1018).
Output
Print the required number.
Sample Input
Input
1 1 10
Output
10
Input
2 -4 4
Output
5
FAQ | About Virtual Judge | Forum | Discuss | Open
Source Project
All Copyright Reserved ©2010-2014 HUST ACM/ICPC TEAM
Anything about the OJ, please ask in the forum, or contact author:Isun, but
NEVER contact freefcw@gmail.com, who is irrelevant to this site !!
Server Time:
姐:两种情况!
1.b<0, b如果不能整除以k 。则产生物差1.
2.a>0 a如果不能整除以k。则产生误差1.
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int main()
{
long long a,b,n,i,j,l,k,t,s;
while(scanf("%lld%lld%lld",&k,&a,&b)!=EOF)
{
s=b/k-a/k+1;
if(a>=0)
{
if(a%k!=0)
s--;
}
else
if(b<0&&b%k!=0)
s--;
printf("%lld\n",s);
}
return 0;
}
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