标签:esc any dex cstring ted 等于 desc put 尺取法
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.
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).
For each test case, output a single integer in a line -- the longest straight ZB can get.
你有n张牌在1~~m之间取值,0可以表示成任意一张牌(也就是王牌),
现在问你你手里的牌最长有几张牌是连续的。
思路:
记录卡牌的值,和零的值,求出1至M之间缺少的个数是否小于等于零的个数,求出最大值,
队友竟然用了尺取法!
代码
#include<iostream> #include<algorithm> #include<cstring> #include<cmath> using namespace std; int main() { std::ios::sync_with_stdio(false); int n,m,t,i,j,l,r,a,num[100001],b[100001],ans; cin>>t; while(t--) { ans=0; memset(num,0,sizeof(num)); memset(b,0,sizeof(b)); cin>>n>>m; for(i=0;i<n;i++) { cin>>a; num[a]++; } ans=0; for(i=1;i<=m;i++) if(num[i]==0)b[i]=b[i-1]+1;else b[i]=b[i-1]; int l=0,r=0; while(r<=m) { if(b[r]-b[l]<=num[0]) { ans=max(ans,r-l); r++; } else l++; } cout<<ans<<endl; } }
标签:esc any dex cstring ted 等于 desc put 尺取法
原文地址:https://www.cnblogs.com/luowentao/p/8997034.html