CodeForces 558A

题意：给定一些苹果树的位置和树上的苹果数，然后一个人站在原点，每次碰到苹果就往相反的方向走，问能得到的最大苹果数。

思路：直接模拟即可。先假设往左走，然后再假设往右走。遍历一遍即可。

code：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <sstream>
#include <string>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>using namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;const int INF=0x3fffffff;
const int inf=-INF;
const int N=1000000;
const int M=205;
const int mod=1000000007;
const double pi=acos(-1.0);#define cls(x,c) memset(x,c,sizeof(x))
#define cpy(x,a) memcpy(x,a,sizeof(a))
#define ft(i,s,n) for (int i=s;i<=n;i++)
#define frt(i,s,n) for (int i=s;i>=n;i--)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lrt  rt<<1
#define rrt  rt<<1|1
#define middle int m=(r+l)>>1
#define lowbit(x) (x&-x)
#define pii pair<int,int>
#define mk make_pair
#define IN freopen("in.txt","r",stdin);
#define OUT freopen("out.txt","w",stdout);struct node
{int x,a;
}g[M];
bool cmp(node A,node B){return A.x<B.x;
}
int main()
{int n;scanf("%d",&n);ft(i,1,n) scanf("%d %d",&g[i].x,&g[i].a);sort(g+1,g+1+n,cmp);g[0].a=g[0].x=0;g[n+1].a=g[n+1].x=0;int t=1,ans=0,s=0;ft(i,2,n){if (g[i-1].x<0&&g[i].x>0){ t=i;break;}}if (t==1){if (g[1].x>0) printf("%d\n",g[1].a);else printf("%d\n",g[n].a);return 0;}s=g[t].a;for (int p=t-1,q=t+1;;p--,q++){if (g[p].a) s+=g[p].a;else  break;if (g[q].a) s+=g[q].a;else break;}ans=g[t-1].a;for (int p=t-2,q=t;;p--,q++){if (g[q].a) ans+=g[q].a;else break;if (p<0) break;if (g[p].a) ans+=g[p].a;else break;}ans=max(ans,s);printf("%d\n",ans);
}

CodeForces 558B

题意：在不改变整个数组出现最多的数的次数的条件下使区间尽量小。

思路：在输入的时候，记录某个数的左右区间。然后遍历，当某个数的出现次数等于最大值时就缩减。

code：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <sstream>
#include <string>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>using namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;const int INF=0x3fffffff;
const int inf=-INF;
const int N=1000000+5;
const int M=1e5+5;
const int mod=1000000007;
const double pi=acos(-1.0);#define cls(x,c) memset(x,c,sizeof(x))
#define cpy(x,a) memcpy(x,a,sizeof(a))
#define ft(i,s,n) for (int i=s;i<=n;i++)
#define frt(i,s,n) for (int i=s;i>=n;i--)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lrt  rt<<1
#define rrt  rt<<1|1
#define middle int m=(r+l)>>1
#define lowbit(x) (x&-x)
#define pii pair<int,int>
#define mk make_pair
#define IN freopen("in.txt","r",stdin);
#define OUT freopen("out.txt","w",stdout);int n,v[M];
int ct[N];
struct node
{int x,y;
}g[N];
int main()
{scanf("%d",&n);cls(ct,0);ft(i,1,n) {scanf("%d",&v[i]);int t=v[i];if (ct[t]==0)g[t].x=i;g[t].y=i;ct[t]++;}int mx=0;ft(i,1,n) mx=max(mx,ct[v[i]]);int al=1,ar=n;ft(i,1,N){if (ct[i]==mx){int ti=g[i].x,tj=g[i].y;if (tj-ti<ar-al){al=ti;ar=tj;}}}printf("%d %d\n",al,ar);
}