Educational Codeforces Round 46 (Rated for Div. 2)

链接http://codeforces.com/contest/1000

B. Light It Up

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Recently, you bought a brand new smart lamp with programming features. At first, you set up a schedule to the lamp. Every day it will turn power on at moment $\mathrm{00 and\; turn\; power\; off\; at\; moment MM.\; Moreover,\; the\; lamp\; allows\; you\; to\; set\; a\; program\; of\; switching\; its\; state\; (states\; are\; "lights\; on"\; and\; "lights\; off").\; Unfortunately,\; some\; program\; is\; already\; installed\; into\; the\; lamp.}$

The lamp allows only good programs. Good program can be represented as a non-empty array $\mathrm{aa,\; where 0<a1<a2<\cdots <a|a|<M0<a1<a2<\cdots <a|a|<M.\; All aiai must\; be\; integers.\; Of\; course,\; preinstalled\; program\; is\; a\; good\; program.}$

The lamp follows program $\mathrm{aa in\; next\; manner:\; at\; moment 00 turns\; power\; and\; light\; on.\; Then\; at\; moment aiai the\; lamp\; flips\; its\; state\; to\; opposite\; (if\; it\; was\; lit,\; it\; turns\; off,\; and\; vice\; versa).\; The\; state\; of\; the\; lamp\; flips\; instantly:\; for\; example,\; if\; you\; turn\; the\; light\; off\; at\; moment 11 and\; then\; do\; nothing,\; the\; total\; time\; when\; the\; lamp\; is\; lit\; will\; be 11.\; Finally,\; at\; moment MM the\; lamp\; is\; turning\; its\; power\; off\; regardless\; of\; its\; state.}$

Since you are not among those people who read instructions, and you don't understand the language it's written in, you realize (after some testing) the only possible way to alter the preinstalled program. You can insert at most one element into the program $\mathrm{aa,\; so\; it\; still\; should\; be\; a good program\; after\; alteration.\; Insertion\; can\; be\; done\; between\; any\; pair\; of\; consecutive\; elements\; of aa,\; or\; even\; at\; the\; begining\; or\; at\; the\; end\; of aa.}$

Find such a way to alter the program that the total time when the lamp is lit is maximum possible. Maybe you should leave program untouched. If the lamp is lit from $\mathrm{xx till\; moment yy,\; then\; its\; lit\; for y-xy-x units\; of\; time.\; Segments\; of\; time\; when\; the\; lamp\; is\; lit\; are\; summed\; up.}$

Input

First line contains two space separated integers $\mathrm{nn and MM (1\le n\le 1051\le n\le 105, 2\le M\le 1092\le M\le 109)\; —\; the\; length\; of\; program aa and\; the\; moment\; when\; power\; turns\; off.}$

Second line contains $\mathrm{nn space\; separated\; integers a1,a2,\dots ,ana1,a2,\dots ,an (0<a1<a2<\cdots <an<M0<a1<a2<\cdots <an<M)\; —\; initially\; installed\; program aa.}$

Output

Print the only integer — maximum possible total time when the lamp is lit.

Examples

input

Copy

3 10
4 6 7

output

Copy

8

input

Copy

2 12
1 10

output

Copy

9

input

Copy

2 7
3 4

output

Copy

6

Note

In the first example, one of possible optimal solutions is to insert value $\mathrm{x=3x=3 before a1a1,\; so\; program\; will\; be [3,4,6,7][3,4,6,7] and\; time\; of\; lamp\; being\; lit\; equals (3-0)+(6-4)+(10-7)=8(3-0)+(6-4)+(10-7)=8.\; Other\; possible\; solution\; is\; to\; insert x=5x=5 in\; appropriate\; place.}$

In the second example, there is only one optimal solution: to insert $\mathrm{x=2x=2 between a1a1 and a2a2.\; Program\; will\; become [1,2,10][1,2,10],\; and\; answer\; will\; be (1-0)+(10-2)=9(1-0)+(10-2)=9.}$

In the third example, optimal answer is to leave program untouched, so answer will be $(3-0)+(7-4)=6(3-0)+(7-4)=6.$

B. Light It Up

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Recently, you bought a brand new smart lamp with programming features. At first, you set up a schedule to the lamp. Every day it will turn power on at moment $\mathrm{00 and\; turn\; power\; off\; at\; moment MM.\; Moreover,\; the\; lamp\; allows\; you\; to\; set\; a\; program\; of\; switching\; its\; state\; (states\; are\; "lights\; on"\; and\; "lights\; off").\; Unfortunately,\; some\; program\; is\; already\; installed\; into\; the\; lamp.}$

The lamp allows only good programs. Good program can be represented as a non-empty array $\mathrm{aa,\; where 0<a1<a2<\cdots <a|a|<M0<a1<a2<\cdots <a|a|<M.\; All aiai must\; be\; integers.\; Of\; course,\; preinstalled\; program\; is\; a\; good\; program.}$

The lamp follows program $\mathrm{aa in\; next\; manner:\; at\; moment 00 turns\; power\; and\; light\; on.\; Then\; at\; moment aiai the\; lamp\; flips\; its\; state\; to\; opposite\; (if\; it\; was\; lit,\; it\; turns\; off,\; and\; vice\; versa).\; The\; state\; of\; the\; lamp\; flips\; instantly:\; for\; example,\; if\; you\; turn\; the\; light\; off\; at\; moment 11 and\; then\; do\; nothing,\; the\; total\; time\; when\; the\; lamp\; is\; lit\; will\; be 11.\; Finally,\; at\; moment MM the\; lamp\; is\; turning\; its\; power\; off\; regardless\; of\; its\; state.}$

Since you are not among those people who read instructions, and you don't understand the language it's written in, you realize (after some testing) the only possible way to alter the preinstalled program. You can insert at most one element into the program $\mathrm{aa,\; so\; it\; still\; should\; be\; a good program\; after\; alteration.\; Insertion\; can\; be\; done\; between\; any\; pair\; of\; consecutive\; elements\; of aa,\; or\; even\; at\; the\; begining\; or\; at\; the\; end\; of aa.}$

Find such a way to alter the program that the total time when the lamp is lit is maximum possible. Maybe you should leave program untouched. If the lamp is lit from $\mathrm{xx till\; moment yy,\; then\; its\; lit\; for y-xy-x units\; of\; time.\; Segments\; of\; time\; when\; the\; lamp\; is\; lit\; are\; summed\; up.}$

Input

First line contains two space separated integers $\mathrm{nn and MM (1\le n\le 1051\le n\le 105, 2\le M\le 1092\le M\le 109)\; —\; the\; length\; of\; program aa and\; the\; moment\; when\; power\; turns\; off.}$

Second line contains $\mathrm{nn space\; separated\; integers a1,a2,\dots ,ana1,a2,\dots ,an (0<a1<a2<\cdots <an<M0<a1<a2<\cdots <an<M)\; —\; initially\; installed\; program aa.}$

Output

Print the only integer — maximum possible total time when the lamp is lit.

Examples

input

Copy

3 10
4 6 7

output

Copy

8

input

Copy

2 12
1 10

output

Copy

9

input

Copy

2 7
3 4

output

Copy

6

Note

In the first example, one of possible optimal solutions is to insert value $\mathrm{x=3x=3 before a1a1,\; so\; program\; will\; be [3,4,6,7][3,4,6,7] and\; time\; of\; lamp\; being\; lit\; equals (3-0)+(6-4)+(10-7)=8(3-0)+(6-4)+(10-7)=8.\; Other\; possible\; solution\; is\; to\; insert x=5x=5 in\; appropriate\; place.}$

In the second example, there is only one optimal solution: to insert $\mathrm{x=2x=2 between a1a1 and a2a2.\; Program\; will\; become [1,2,10][1,2,10],\; and\; answer\; will\; be (1-0)+(10-2)=9(1-0)+(10-2)=9.}$

In the third example, optimal answer is to leave program untouched, so answer will be $(3-0)+(7-4)=6(3-0)+(7-4)=6.$

$这是一道用贪心算法的题目，即在区间中插入一个数，整个序列最多能增加多少。假设区间为[a,b]，则插入的值为a+1或b-1，取决于插入的位置是开还是关。$

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<string>
#include<vector>
#define DEBUG(x) cout<<#x<<" = "<<x<<endl
using namespace std;
typedef long long ll;
const int MAXN=1e5+10;
int n,M;
int prog[MAXN];
ll litTime[MAXN];
ll OnLitTimeToEnd[MAXN];
ll OffLitTimeToEnd[MAXN];
int main()
{
//    freopen("in.txt","r",stdin);
    scanf("%d%d",&n,&M);
    prog[0]=0,prog[n+1]=M;
    litTime[0]=0;
    for(int i=1;i<=n+1 ;i++ ){
        if(i<n+1)
            scanf("%d",&prog[i]);
        if(i%2)litTime[i]=litTime[i-1]+prog[i]-prog[i-1];
        else litTime[i]=litTime[i-1];
    }
    OffLitTimeToEnd[n+1]=0;
    OnLitTimeToEnd[n+1]=0;
    for(int i=n;i>=0 ;i-- ){
        OnLitTimeToEnd[i]=OffLitTimeToEnd[i+1]+prog[i+1]-prog[i];
///踩了一个坑     OnLitTimeToEnd[i]=OffLitTimeToEnd[i-1]+prog[i+1]-prog[i];
        OffLitTimeToEnd[i]=OnLitTimeToEnd[i+1];
    }
    int ans=litTime[n+1];
    for(int i=1;i<=n+1 ;i++ ){
        int t;
        if(i%2){///关
            t=prog[i]-1-prog[i-1]+litTime[i-1]+OnLitTimeToEnd[i];
        }
        else {///开
            t=prog[i]-(prog[i-1]+1)+litTime[i-1]+OffLitTimeToEnd[i];
        }
        ans=max(ans,t);
    }
    printf("%d
",ans);
}

C. Covered Points Count

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given $\mathrm{nn segments\; on\; a\; coordinate\; line;\; each\; endpoint\; of\; every\; segment\; has\; integer\; coordinates.\; Some\; segments\; can\; degenerate\; to\; points.\; Segments\; can\; intersect\; with\; each\; other,\; be\; nested\; in\; each\; other\; or\; even\; coincide.}$

Your task is the following: for every $\mathrm{k\in [1..n]k\in [1..n],\; calculate\; the\; number\; of\; points\; with\; integer\; coordinates\; such\; that\; the\; number\; of\; segments\; that\; cover\; these\; points\; equals kk.\; A\; segment\; with\; endpoints lili and riri covers\; point xx if\; and\; only\; if li\le x\le rili\le x\le ri.}$

Input

The first line of the input contains one integer $\mathrm{nn (1\le n\le 2\cdot 1051\le n\le 2\cdot 105)\; —\; the\; number\; of\; segments.}$

The next $\mathrm{nn lines\; contain\; segments.\; The ii-th\; line\; contains\; a\; pair\; of\; integers li,rili,ri (0\le li\le ri\le 10180\le li\le ri\le 1018)\; —\; the\; endpoints\; of\; the ii-th\; segment.}$

Output

Print $\mathrm{nn space\; separated\; integers cnt1,cnt2,\dots ,cntncnt1,cnt2,\dots ,cntn,\; where cnticnti is\; equal\; to\; the\; number\; of\; points\; such\; that\; the\; number\; of\; segments\; that\; cover\; these\; points\; equals\; to ii.}$

Examples

input

Copy

3
0 3
1 3
3 8

output

Copy

6 2 1 

input

Copy

3
1 3
2 4
5 7

output

Copy

5 2 0 

Note

The picture describing the first example:

Points with coordinates $[0,4,5,6,7,8][0,4,5,6,7,8] are\; covered\; by\; one\; segment,\; points [1,2][1,2] are\; covered\; by\; two\; segments\; and\; point [3][3] is\; covered\; by\; three\; segments.$

The picture describing the second example:

Points $[1,4,5,6,7][1,4,5,6,7] are\; covered\; by\; one\; segment,\; points [2,3][2,3] are\; covered\; by\; two\; segments\; and\; there\; are\; no\; points\; covered\; by\; three\; segments.$

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<string>
#include<vector>
#define DEBUG(x) cout<<#x<<" = "<<x<<endl
typedef long long ll;
using namespace std;
const int MAXN=4e5+10;
struct Point{
    ll pos;
    ll delta;
    Point(){}
    Point(ll p,ll d):pos(p),delta(d){}
    bool operator<(const Point &p)const
    {
        if(pos!=p.pos)return pos<p.pos;
        return delta>p.delta;
    }
};
int N;
Point points[MAXN];
ll ans[MAXN];
int main()
{
//    freopen("in.txt","r",stdin);
    scanf("%d",&N);
    for(int i=0;i<2*N ;i++ ){
        scanf("%lld",&points[i].pos);
        if(i%2==0)points[i].delta=1;
        else {
                points[i].delta=-1;
                ///需要包括右端点
                points[i].pos++;
        }
    }
    sort(points,points+2*N);
    int curCov=0;
    for(int i=0;i<2*N-1 ;i++ ){
        curCov+=points[i].delta;
        ans[curCov]+=points[i+1].pos-points[i].pos;
    }
    for(int i=1;i<=N ;i++ ){
        printf("%lld",ans[i]);
        if(i==N)printf("
");
        else printf(" ");
    }
}