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Testing Round #12 A,B,C 讨论，贪心，树状数组优化dp
题目链接：http://codeforces.com/contest/597
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.
The only line contains three space-separated integers k, a and b (1 ≤ k ≤ 1018; - 1018 ≤ a ≤ b ≤ 1018).
Print the required number.
1 1 10
10
2 -4 4
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;}
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.
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).
Print the maximal number of orders that can be accepted.
2
7 11
4 7
1
5
1 2
2 3
3 4
4 5
5 6
3
6
4 8
1 5
4 7
2 5
1 3
6 8
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;}
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.
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.
Print one integer — the answer to the problem.
5 2
1
2
3
5
4
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;}
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.
The only line contains three space-separated integers k, a and b (1 ≤ k ≤ 1018; - 1018 ≤ a ≤ b ≤ 1018).
Print the required number.
1 1 10
10
2 -4 4
5
Testing Round #12 A,B,C 讨论，贪心，树状数组优化dp