hdu5773The All-purpose Zero,1257最少拦截次数

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这次的两题都是求连续的递增序列，我主要是用求最长连续子序列的方法解决的，核心（int po=lower_bound(ans,ans+len,b[i])-ans;），理解这句代码，节能解决一类题

5773The All-purpose Zero

#include<cstring>#include<cstdio>#include<algorithm>using namespace std;int a[100001],ans[100001],b[100001];int main(){    int i,t,n,zero,count,len,h=1,flag;    scanf("%d",&t);    while(t--)        {        zero=0;        count=0;        flag=1;        scanf("%d",&n);        for(i=1;i<=n;i++)        {            scanf("%d",&a[i]);        }        memset(b,0,sizeof(b));        memset(ans,0,sizeof(ans));        for(i=1;i<=n;i++)        {            if(a[i]==0) zero++;            else             {                flag=0;                b[count++]=a[i]-zero;            }        }        ans[0]=b[0];        len=1;        for(i=1;i<count;i++)        {            if(ans[len-1]<b[i]) ans[len++]=b[i];            else             {                int po=lower_bound(ans,ans+len,b[i])-ans;                ans[po]=b[i];            }            }        if(flag==1) len=0;        printf("Case #%d: %d\n",h++,len+zero);    }    return 0;}

1257最少拦截次数

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int a[100001],b[100001];int main(){    int n,i,count;    while(scanf("%d",&n)!=EOF)    {        count=0;        for(i=1;i<=n;i++)        {            scanf("%d",&a[i]);        }        b[0]=a[0];        int len=1;        for(i=1;i<=n;i++)        {            if(b[len-1]<a[i]) b[len++]=a[i];            else             {                int pos = lower_bound(b,b+len,a[i])-b;                b[pos] = a[i];            }        }                         printf("%d\n",len-1);    }  }

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