2017 多校赛 第二场

1003.Maximum Sequence

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0

Problem Description

Steph is extremely obsessed with “sequence problems” that are usually seen on magazines: Given the sequence 11, 23, 30, 35, what is the next number? Steph always finds them too easy for such a genius like himself until one day Klay comes up with a problem and ask him about it.

Given two integer sequences {ai} and {bi} with the same length n, you are to find the next n numbers of {ai}: an+1…a2n. Just like always, there are some restrictions on an+1…a2n: for each number ai, you must choose a number bk from {bi}, and it must satisfy ai≤max{aj-j│bk≤j<i}, and any bk can’t be chosen more than once. Apparently, there are a great many possibilities, so you are required to find max{∑2nn+1ai} modulo 109+7 .

Now Steph finds it too hard to solve the problem, please help him.

 

Input

The input contains no more than 20 test cases.
For each test case, the first line consists of one integer n. The next line consists of n integers representing {ai}. And the third line consists of n integers representing {bi}.
1≤n≤250000, n≤a_i≤1500000, 1≤b_i≤n.

 

Output

For each test case, print the answer on one line: max{∑2nn+1ai} modulo 109+7。

 

Sample Input

4 8 11 8 5 3 1 4 2

 

Sample Output

27

 

思路：贪心，用一个数组记下从i到n的a[i]-i的最大值pre[i],可以得到pre[i]=max(pre[i+1],a[i]-i)，然后容易得到b[k]从小往大取时，后n项之和最大，所以a[n+1]=pre[b[1]]（后n项会呈单调不递增趋势，a[n+1]-n-1即为之后取最大值的比较对象）。

代码：

#include "cstdio"
#include "stdlib.h"
#include "iostream"
#include "algorithm"
#include "string"
#include "cstring"
#include "queue"
#include "cmath"
#include "vector"
#include "map"
#include "set"
#define db double
#define ll long long
#define inf 0x3f3f3f
using namespace std;
const int N=3e5+5;
const int mod=1e9+7;
#define rep(i,x,y) for(int i=x;i<=y;i++)
//char s[N],t[N];
db  pi=3.14;
//int s[N],w[N];
int a[N],b[N];
int pre[N];
int main()
{
    int n;
    while(scanf("%d",&n)==1){
        for(int i=1;i<=n;i++){
            scanf("%d",a+i);
            a[i]-=i;
        }
        memset(pre,0, sizeof(pre));
        for(int i=n;i>=1;i--) pre[i]=max(a[i],pre[i+1]);
        for(int i=1;i<=n;i++) scanf("%d",&b[i]);
        sort(b+1,b+n+1);
        int ma=0;
        ll ans=pre[b[1]];
        for(int i=2;i<=n;i++){
            ma=max(pre[b[1]]-n-1,pre[b[i]]);
            ans=(ma+ans)%mod;
//            printf("%d
",ma);
        }
        printf("%lld
",ans);
    }

}

10011.Regular polygon

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 752    Accepted Submission(s): 273

Problem Description

On a two-dimensional plane, give you n integer points. Your task is to figure out how many different regular polygon these points can make.

 

Input

The input file consists of several test cases. Each case the first line is a numbers N (N <= 500). The next N lines ,each line contain two number Xi and Yi(-100 <= xi,yi <= 100), means the points’ position.(the data assures no two points share the same position.)

 

Output

For each case, output a number means how many different regular polygon these points can make.

 

Sample Input

4

0 0

0 1

1 0

1 1

6

0 0

0 1

1 0

1 1

2 0

2 1

 

Sample Output

1 2

思路：整点正多边形只能是正方形，将点按x从小到大，y从大到小，枚举对角线 (x1,y1) (x2,y2) xx=x1-x2,yy=y1-y2;则x3=x1+yy,y3=y1-xx;x4=x1-yy,y4=y1+xx;判断是否存在即可。

学长的代码：

#include "cstdio"
#include "stdlib.h"
#include "iostream"
#include "algorithm"
#include "string"
#include "cstring"
#include "queue"
#include "cmath"
#include "vector"
#include "map"
#include "set"
#define db double
#define ll long long
#define inf 0x3f3f3f
#define rep(i,x,y) for(int i=x;i<=y;i++)
using namespace std;
const int N=3e5+5;
const int mod=1e9+7;
const double eps=1e-9;
//char s[N],t[N];
db  pi=3.14;
int n;
struct node
{
    double x,y;
}p[N];
bool cmp(node a,node b)
{
    return (a.x<b.x||(a.x==b.x&&a.y<b.y));
}
bool Check(double x,double y)
{
    int l=1,r=n;
    while(l<=r)
    {
        int mid=l+r>>1;
        if(fabs(p[mid].x-x)<eps&&fabs(p[mid].y-y)<eps) return true;
        else if(p[mid].x-x>eps||(fabs(p[mid].x-x)<eps&&p[mid].y-y>eps)) r=mid-1;
        else l=mid+1;
    }
    return false;
}
int main()
{
    while(~scanf("%d",&n))
    {
        ll ans=0;
        rep(i,1,n)
        {
            scanf("%lf%lf",&p[i].x,&p[i].y);
        }
        sort(p+1,p+1+n,cmp);
        rep(i,1,n)
        {
            rep(j,i+1,n)
            {
                double x=(p[i].x+p[j].x)/2;
                double y=(p[i].y+p[j].y)/2;
                double xx=p[i].x-x;
                double yy=p[i].y-y;
                if(Check(x+yy,y-xx)&&Check(x-yy,y+xx)) ans++;
            }
        }
        printf("%lld
",ans/2);
    }
}