FZUoj 题目2216 The Longest Straight*（二分）

来源：互联网发布：淘宝客发单机器人编辑：程序博客网时间：2024/06/05 06:42

Problem 2216 The Longest Straight

Accept: 22 Submit: 44
Time Limit: 1000 mSec Memory Limit : 32768 KB

Problem Description

ZB is playing a card game where the goal is to make straights. Each card in the deck has a number between 1 and M(including 1 and M). A straight is a sequence of cards with consecutive values. Values do not wrap around, so 1 does not come after M. In addition to regular cards, the deck also contains jokers. Each joker can be used as any valid number (between 1 and M, including 1 and M).

You will be given N integers card[1] .. card[n] referring to the cards in your hand. Jokers are represented by zeros, and other cards are represented by their values. ZB wants to know the number of cards in the longest straight that can be formed using one or more cards from his hand.

Input

The first line contains an integer T, meaning the number of the cases.

For each test case:

The first line there are two integers N and M in the first line (1 <= N, M <= 100000), and the second line contains N integers card[i] (0 <= card[i] <= M).

Output

For each test case, output a single integer in a line -- the longest straight ZB can get.

Sample Input

27 110 6 5 3 0 10 118 1000100 100 100 101 100 99 97 103

Sample Output

53

Source

第六届福建省大学生程序设计竞赛-重现赛（感谢承办方华侨大学）

题目大意：一个n，m,后边n个数，0的个数代表鬼牌，龟牌什么都可以当，问选出的最大的连续的值得长度是多少

思路：暴力左端点二分右端点，，

ac代码

#include<stdio.h>#include<string.h>#include<stdlib.h>#include<iostream>#include<math.h>#include<algorithm>#define LL long long#define INF 0x3f3f3f3fusing namespace std;int vis[100010],num[100010];void init(int m){    int i;    for(i=0;i<=m;i++)        vis[i]=0;}int bseach_l(int l,int r,int val){    int ans=r;    int tl=l;    while(l<=r)    {        int mid=(l+r)>>1;        if(num[mid]-num[tl]>val)        {           // ans=mid;            //l=mid+1;            r=mid-1;        }        else            l=mid+1;    }    return r;}int main(){    int t;    scanf("%d",&t);    while(t--)    {        int n,m;        scanf("%d%d",&n,&m);        int i;        int cnt=0;        memset(vis,0,sizeof(vis));        for(i=0;i<n;i++)        {            int x;            scanf("%d",&x);            if(x==0)            {                cnt++;                continue;            }            vis[x]=1;        }        int ans=0,tc=cnt,pt;        memset(num,0,sizeof(num));        for(i=1;i<=m;i++)        {            if(vis[i])                num[i]=num[i-1];            else                num[i]=num[i-1]+1;        }        for(i=0;i<=m;i++)        {            int temp=bseach_l(i,m,cnt);            ans=max(temp-i,ans);        }        printf("%d\n",ans);    }}

0 0