hdu 3 * problem

hdu 6182

给出 $n$

求 $\sum_{i = 1} ^ {\infty} (i * i <= n)$

暴力枚举

hdu 6186

给出 $n$ 个数

$1e6$ 次询问，每次询问这 $n$ 个数不包含第 $p$ 个时的 $xor, or and$ 的值

前缀 + 后缀处理

hdu 6188

给出 $n$ 个数$A_i$

$a.$ 每两个相同的数对答案的贡献为 $1$

$b.$ 每相连的 $3$ 个数对答案的贡献为 $1$

每个数只能使用一次

最大化答案

贪心：取 $i，i - 1, i - 2$ 这个顺子的时候先把 $i - 1，i - 2$ 的对子取掉，取完再取 $i$ 的对子

Code

6182

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>

using namespace std;

long long Ksm(long long a, long long b) {
    long long ret = ;
    while(b) {
        if(b & ) ret = ret * a;
        a = a * a;
        b >>= ;
    }
    return ret;
}

#define LL long long

LL num[];

int main() {
    for(LL i = ; i <= ; i ++) num[i] = Ksm(i, i);
    LL a;
    while(cin >> a) {
        if(a == ) {
            cout <<  << "\n";
            continue;
        }
        int i;
        for(i = ; i <= ; i ++) {
            if(num[i] > a) break;
        }
        cout << i -  << "\n";
    }

    return ;
} 

6186

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>

using namespace std;

#define LL long long

const int N = 1e5 + ;
int A[N];
int And[N], Or[N], Xor[N];
int And2[N], Or2[N], Xor2[N];
int n, m;

#define gc getchar()

inline int read() {
    int x = ; char c = gc;
    while(c < '' || c > '') c = gc;
    while(c >= '' && c <= '') x = x *  + c - '', c = gc;
    return x;
}

int main() {
    while(scanf("%d%d", &n, &m) == ) {
        And[] = And2[n + ] = ~ ;
        Xor[] = Xor2[n + ] = ;
        Or[] = Or2[n + ] = ;
        for(int i = ; i <= n; i ++) A[i] = read(); //cin >> A[i];
        for(int i = ; i <= n; i ++) And[i] = (And[i - ] & A[i]);
        for(int i = ; i <= n; i ++) Or[i] = (Or[i - ] | A[i]);
        for(int i = ; i <= n; i ++) Xor[i] = (Xor[i - ] ^ A[i]);
        for(int i = n; i >= ; i --) And2[i] = (And2[i + ] & A[i]);
        for(int i = n; i >= ; i --) Or2[i] = (Or2[i + ] | A[i]);
        for(int i = n; i >= ; i --) Xor2[i] = (Xor2[i + ] ^ A[i]);
        for(; m; m --) {
            int p = read();
            printf("%d ", (And[p - ] & And2[p + ]));
            printf("%d ", (Or[p - ] | Or2[p + ]));
            printf("%d\n", (Xor[p - ] ^ Xor2[p + ]));
        }
    }

    return ;
} 

6188

#include <iostream>
#include <cstdio>
#include <cstring>

const int N = 1e6 + ;

#define gc getchar()
inline int read() {
    int x = ; char c = gc;
    while(c < '' || c > '') c = gc;
    while(c >= '' && c <= '') x = x *  + c - '', c = gc;
    return x;
}

int n;
int T[N];

int main() {
    while(~ scanf("%d", &n)) {
        memset(T, , sizeof T);
        int Answer = ;
        for(int i = ; i <= n; i ++) T[read()] ++;
        for(int i = ; i <= (int)1e6; i ++) {
            if(i >=  && T[i] && T[i - ] && T[i - ]) {
                Answer ++;
                T[i] --, T[i - ] --, T[i - ] --;
            }
            Answer += (T[i] >> );
            T[i] %= ;
        }
        printf("%d\n", Answer);
    }

    return ;
}