题目链接：http://codeforces.com/contest/597
A. Divisibility
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
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 that a ≤ 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.
Examples
input
1 1 10
output
10
input
2 -4 4
output
5
题意：找出[a,b]区间内整除k的数的个数；
思路:小心点特判即可；
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-14
#define bug(x) cout<<"bug"<<x<<endl;
const int N=1e5+10,M=4e6+10,inf=2147483647;
const ll INF=1e18+10,mod=1e9+7;
/// 数组大小
int main()
{
ll k,a,b;
scanf("%lld%lld%lld",&k,&a,&b);
if(a<=0&&b>=0)
printf("%lld\n",(b/k)-(a/k)+1);
else if(a>=0&&b>=0)
printf("%lld\n",(b/k)-(a/k+(a%k?1:0))+1);
else
printf("%lld\n",(abs(a)/k)-(abs(b)/k+(abs(b)%k?1:0))+1);
return 0;
}
B. Restaurant
time limit per test
4 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
A restaurant received n orders for the rental. Each rental order reserve the restaurant for a continuous period of time, the i-th order is characterized by two time values — the start time li and the finish time ri (li ≤ ri).
Restaurant management can accept and reject orders. What is the maximal number of orders the restaurant can accept?
No two accepted orders can intersect, i.e. they can‘t share even a moment of time. If one order ends in the moment other starts, they can‘t be accepted both.
Input
The first line contains integer number n (1 ≤ n ≤ 5·105) — number of orders. The following n lines contain integer values li and ri each (1 ≤ li ≤ ri ≤ 109).
Output
Print the maximal number of orders that can be accepted.
Examples
input
27 114 7
output
1
input
51 22 33 44 55 6
output
3
input
64 81 54 72 51 36 8
output
2
贪心：按r从小到大排序即可；
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-14
#define bug(x) cout<<"bug"<<x<<endl;
const int N=5e5+10,M=4e6+10,inf=2147483647;
const ll INF=1e18+10,mod=1e9+7;
struct is
{
int l,r;
bool operator <(const is &c)const
{
return r<c.r;
}
}a[N];
/// 数组大小
int main()
{
int n;
scanf("%d",&n);
for(int i=1;i<=n;i++)
scanf("%d%d",&a[i].l,&a[i].r);
sort(a+1,a+1+n);
int s=0,ans=0;
for(int i=1;i<=n;i++)
{
if(a[i].l>s)
{
ans++;
s=a[i].r;
}
}
printf("%d\n",ans);
return 0;
}
C. Subsequences
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
For the given sequence with n different elements find the number of increasing subsequences with k + 1 elements. It is guaranteed that the answer is not greater than 8·1018.
Input
First line contain two integer values n and k (1 ≤ n ≤ 105, 0 ≤ k ≤ 10) — the length of sequence and the number of elements in increasing subsequences.
Next n lines contains one integer ai (1 ≤ ai ≤ n) each — elements of sequence. All values ai are different.
Output
Print one integer — the answer to the problem.
Examples
input
5 212354
output
7
题意：给你n个数，最长上升子序列长度为k+1的个数；
思路：看下数据范围k<10很关键,dp[i][j]表示以i为结束长度为j时候的方案数
现在你到i的时候你只需要再T[j]（树状数组)的a[i]的位置表示方案数；
统计小于a[i]的方案，k==0时候特判；
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-14
#define bug(x) cout<<"bug"<<x<<endl;
const int N=1e5+10,M=4e6+10,inf=2147483647;
const ll INF=1e18+10,mod=1e9+7;
struct AYT
{
ll tree[N];
int lowbit(int x)
{
return x&-x;
}
void update(int x,ll c)
{
while(x<N)
{
tree[x]+=c;
x+=lowbit(x);
}
}
ll query(int x)
{
ll ans=0;
while(x)
{
ans+=tree[x];
x-=lowbit(x);
}
return ans;
}
};
AYT T[12];
/// 数组大小
int a[N];
int main()
{
int n,k;
scanf("%d%d",&n,&k);
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
if(k==0)
return 0*printf("%d\n",n);
ll ans=0;
for(int i=1;i<=n;i++)
{
ans+=T[k].query(a[i]-1);
for(int j=k;j>=2;j--)
{
ll x=T[j-1].query(a[i]-1);
T[j].update(a[i],x);
}
T[1].update(a[i],1);
}
printf("%lld\n",ans);
return 0;
}
A. Divisibility
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
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 that a ≤ 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.
Examples
input
1 1 10
output
10
input
2 -4 4
output
5