老年选手做点题

NO.1 CodeForces814C

题意：

略；

思路：

一开始觉得2s，3e8是不是莽的过啊~（天真！可笑！）看来我是太爱尺取了！
预处理复杂度直接就减少了一个数量级，预处理num[ c ] [ rp ] （字符 c 在能替换 rp 数量个情况下的最长区间）复杂度O(26*N^2 + q);CC

CODE:

#include 
using namespace std;

#define din(var) scanf("%d",&var)
#define dout(var) printf("%d\n",var);
#define dinll(var) scanf("%lld",&var)
#define doutll(var) printf("%lld\n",var);
#define rep(i,a,n) for (int i=a;i
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define p_b push_back
#define m_p make_pair
#define aLL(x) (x).begin(),(x).end()
#define fi first
#define se second
#define mem(a, b) memset(a, b, sizeof(a))
typedef vector<int> VI;
typedef long double LD;
typedef long long LL;
typedef vector<long long> VIL;

template <class T>
inline bool scan_d(T &ret)
{
    char c;
    int sgn;
    if(c=getchar(),c==EOF) return 0; //EOF
    while(c!='-'&&(c<'0'||c>'9')) c=getchar();
    sgn=(c=='-')?-1:1;
    ret=(c=='-')?0:(c-'0');
    while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0');
    ret*=sgn;
    return 1;
}
inline void out(LL x)
{
    if(x>9) out(x/10);
    putchar(x%10+'0');
}
inline void out(int x)
{
    if(x>9) out(x/10);
    putchar(x%10+'0');
}

inline int Mmax(int x, int y){
    return x > y ? x : y;
}

int num[30][1507], n;
char s[1507];

int main(){
    int len;
    int q, m;
    din(n);
    scanf("%s", s+1);
    rep(c, 0, 26){
        rep(i, 1, n+1){
            int rp = 0;
            rep(j, i, n+1){
                if((s[j]-'a') != c) rp++;
                num[c][rp] = Mmax(num[c][rp], j-i+1);
            }
        }
        rep(i, 1, n+1) num[c][i] = Mmax(num[c][i], num[c][i-1]);
    }
    din(q);
    char op[5];
    while(q--){
        scan_d(m);
        scanf("%s", op);
        dout(num[op[0]-'a'][m]);
    }
    return 0;
}

NO.2 CodeForces811C

题意

段中的所有元素种类的个数=数组的所有元素种类的个数
选几个段，求最大的每个段的异或和；

思路：

DP[ Now ] = DP[ Pre ] + Val{[Pre, Now]};

CODE:

#include 
using namespace std;

#define din(var) scanf("%d",&var)
#define dout(var) printf("%d\n",var);
#define dinll(var) scanf("%lld",&var)
#define doutll(var) printf("%lld\n",var);
#define rep(i,a,n) for (int i=a;i
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define p_b push_back
#define m_p make_pair
#define aLL(x) (x).begin(),(x).end()
#define fi first
#define se second
#define mem(a, b) memset(a, b, sizeof(a))
typedef vector<int> VI;
typedef long double LD;
typedef long long LL;
typedef vector<long long> VIL;

template <class T>
inline bool scan_d(T &ret)
{
    char c;
    int sgn;
    if(c=getchar(),c==EOF) return 0; //EOF
    while(c!='-'&&(c<'0'||c>'9')) c=getchar();
    sgn=(c=='-')?-1:1;
    ret=(c=='-')?0:(c-'0');
    while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0');
    ret*=sgn;
    return 1;
}
inline void out(LL x)
{
    if(x>9) out(x/10);
    putchar(x%10+'0');
}
inline void out(int x)
{
    if(x>9) out(x/10);
    putchar(x%10+'0');
}

inline int Mmax(int x, int y){return x > y ? x : y;}
inline LL Mmax(LL x, LL y){return x > y ? x : y;}
inline int Mmin(int x, int y){return x < y ? x : y;}
inline LL Mmin(LL x, LL y){return x < y ? x : y;}

const int Maxn = 5007;
int dp[Maxn], a[Maxn];
int Left[Maxn], Right[Maxn], n, vis[Maxn];

int main(){
    din(n);
    rep(i, 0, 5001) Left[i] = n, Right[i] = 1;
    rep(i, 1, n+1) din(a[i]), Right[a[i]] = Mmax(Right[a[i]], i), Left[a[i]] = Mmin(Left[a[i]], i);
    rep(i, 1, n+1){
        dp[i] = dp[i-1];
        mem(vis, 0);
        int st = i;
        int ans = 0;
        for(int j=i;j>0;j--){
            if(!vis[a[j]]){
                if(Right[a[j]] > i) break;
                if(Left[a[j]] < st) st = Left[a[j]];
                ans = ans ^ a[j];
                vis[a[j]] = 1;
            }
            if(j <= st) dp[i] = Mmax(dp[i], dp[j-1] + ans);
        }
    }
    dout(dp[n]);
    return 0;
}