Accept: 17 Submit: 39
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
Sample Output
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）
解题思路：首先明确，这几个0是应该连续填放在空里，这样能连成最长的。我们首先记录哪些牌是有的，哪些没有。然后我们枚举左端点，然后通过二分去查找需要0的个数小于总的0的个数的最长的右端点。 mlogm的复杂度。
#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<iostream>
using namespace std;
const int maxn = 1e6+200;
int card[maxn],presum[maxn];
int BinSearch(int l,int r, int k,int key){
int mi = (l+r)/2;
while(l < r){
mi = (l+r)/2;
if(presum[mi]-k > key){
r = mi;
}else{
l = mi+1;
}
}
if(presum[l]-k > key)
l--;
return l;
}
int main(){
int T,n,m;
scanf("%d",&T);
while(T--){
int a, Zero = 0;
scanf("%d%d",&n,&m);
memset(card,0,sizeof(card));
for(int i = 1; i <= n; i++){
scanf("%d",&a);
card[a] = 1;
if(!a) Zero++;
}
for(int i = 1; i <= m; i++){
presum[i] = presum[i-1] + (!card[i]);
}
int ans = 1;
for(int i = 1; i <= m; i++){
int idx = BinSearch(i,m,presum[i-1],Zero);
// printf("%d ",idx);
ans = max(ans, idx-i+1);
}
printf("%d\n",ans);
}
return 0;
} /* 55
11 15
0 0 1 2 4 5 8 9 11 13 14
*/