Codeforces Round #193 (Div. 2)

来源：互联网发布：java zcms 编辑：程序博客网时间：2024/06/07 06:55

题目地址： http://codeforces.com/contest/332

第一题：题目又臭又长，读了好长时间才读懂。

n个人，你是0号，从0开始到n-1循环做动作，只要你前面三个人动作一样，你就喝一杯橙汁，问你能喝多少杯，

模拟

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <cstdlib>#include <cmath>#include <vector>#include <list>#include <deque>#include <queue>#include <iterator>#include <stack>#include <map>#include <set>#include <algorithm>#include <cctype>using namespace std;typedef __int64 LL;const int N=20005;const LL II=1000000007;const int INF=0x3f3f3f3f;const double PI=acos(-1.0);char s[N];int main(){    int i,j,n;    while(cin>>n)    {        scanf("%s",s);        int len=strlen(s),sum=0;        for(i=0;i<len;)        {            if((i+n)<len)            {                i=i+n;                if(s[i-1]==s[i-2]&&s[i-2]==s[i-3])                {                    sum++;                }            }            else                break;        }        cout<<sum<<endl;    }    return 0;}

第二题：给你n个数，要你求一个k使得[a; a + k - 1] 与[b; b + k - 1]的和最大，保证这两个集合不想交。

1、相当于两个dp吧

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <cstdlib>#include <cmath>#include <vector>#include <list>#include <deque>#include <queue>#include <iterator>#include <stack>#include <map>#include <set>#include <algorithm>#include <cctype>using namespace std;typedef __int64 LL;const int N=200005;const LL II=1000000007;const int INF=0x3f3f3f3f;const double PI=acos(-1.0);LL x[N];LL dp[N];int main(){    LL i,j,n,k;    while(scanf("%I64d%I64d",&n,&k)!=EOF)    {        for(i=1;i<=n;i++)            scanf("%I64d",&x[i]);        memset(dp,0,sizeof(dp));        for(i=1;i<=k;i++)            dp[1]+=x[i];        for(i=2;i<=n-k+1;i++)            dp[i]=dp[i-1]-x[i-1]+x[i+k-1];        LL max1=dp[1],Max=0;        int t1,t2,te=1;        for(i=1+k;i<=n-k+1;i++)        {            if((dp[i]+max1)>Max)            {                Max=dp[i]+max1;                t1=te; t2=i;            }            if(max1<dp[i-k+1])            {                max1=dp[i-k+1];                te=i-k+1;            }        }        printf("%d %d\n",t1,t2);    }    return 0;}

2、RMQ算法

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <cstdlib>#include <cmath>#include <vector>#include <list>#include <deque>#include <queue>#include <iterator>#include <stack>#include <map>#include <set>#include <algorithm>#include <cctype>using namespace std;typedef __int64 LL;const int N=200005;const LL II=1000000007;const int INF=0x3f3f3f3f;const double PI=acos(-1.0);LL x[N];LL dp[N];LL d[N][20];LL RMQ(int l,int r){    int k=0;    while((1<<(k+1))<=(r-l+1))        k++;    return max(d[l][k],d[r-(1<<k)+1][k]);}int main(){    int i,j,n,k;    while(scanf("%d%d",&n,&k)!=EOF)    {        for(i=0;i<n;i++)            scanf("%I64d",&x[i]);        memset(dp,0,sizeof(dp));        for(i=0;i<k;i++)            dp[0]+=x[i];        int p=n-k+1;        for(i=1;i<p;i++)            dp[i]=dp[i-1]-x[i-1]+x[i+k-1];        for(i=0;i<p;i++)            d[i][0]=dp[i];        for(j=1;(1<<j)<=p;j++)            for(i=0;i+(1<<j)-1<p;i++)                d[i][j]=max(d[i][j-1],d[i+(1<<(j-1))][j-1]);        LL Max=0,tmp,kk;        int t1,t2;        for(i=0;i<p;i++)        {            if((i+k)<p)            {                LL s=dp[i]+(tmp=RMQ(i+k,p-1));                if(Max<s)                {                    Max=s;  kk=tmp;                    t1=i;   t2=i+k;                }            }        }        for(j=t2;j<p;j++)            if(dp[j]==kk)            {                t2=j;break;            }        printf("%d %d\n",t1+1,t2+1);    }    return 0;}