HDU 4610 cards

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Question: Each number has four evaluations.

1. It is a prime number. 2. The approximate number is a prime number. 3. The sum of an approximate number is a prime number. 4. The product of an approximate number is a complete number.

Select K from N to make the highest score. If the selected K does not meet a certain condition, there is another score.

Http://acm.hdu.edu.cn/showproblem.php? PID = 1, 4610

First, because the data range is not large, 1-1e6 can be preprocessed, And the number used is 20000.

For the first time, the evaluation of all the numbers of 1e6 is completed. The complexity of the direct TLE and N * SQRT (n) is actually only 2 W, and no preprocessing is required, just lay the prime number table.

The first item uses the prime number table directly. After the two or three items are enumerated, the prime number table is used. The product cannot be directly stored in the 4th items before judgment. We can find that when SQRT (n) enumerative approx, each time two divisor I, n/I, their product is N, so the product of divisor is n ^ y * (N is the number of divisor? SQRT (n): 1 ).

Therefore, it can be simplified by side multiplication.

After processing, all numbers can be divided into 2 ^ 4 categories.

The first practice is to enumerate which evaluation types are not selected, and then greedy, wa. after the auction, we find the problem. In this way, a certain evaluation value may not be selected, but the additional value is not added, the extra value may be negative.

The second time: in the case of enumeration 2 ^ 16, it indicates which type to choose. Note that if you enumerate which types, you must obtain at least one. Otherwise, the above situation will still occur. Then you can get greedy.

#include <iostream>#include <cstdio>#include <cstring>#include <vector>#include <cmath>#include <algorithm>using namespace std;typedef long long LL;const int N = 1000001;const int M = 1005;bool isprime[N << 2] = {false}, flag[N][4] = {false} , vis[N] = {false};int score[N] = {0} , state[N];int n , k , b[4] , cnt[1 << 4] , get[1 << 4] , id[1 << 4];bool issquare (LL num) {    LL t = sqrt(num + 0.0);    if(t * t == num || (t + 1) * (t + 1) == num || (t - 1) * (t - 1) == num) return true;    return false;}bool cmp (int a , int b) {    return get[a] > get[b];}void Init() {    for (int i = 0 ; i < N << 2; i ++) {        isprime[i] = true;    }    isprime[1] = false;    for (int i = 2 ; i < N << 2 ; i ++) {        if (! isprime[i]) continue;        for (int j = 2 ; i * j < N << 2 ; j ++) {            isprime[i * j] = false;        }    }    get[0] = 0;    for (int i = 1 ; i < 1 << 4 ; i ++) {        get[i] = get[i >> 1] + (i & 1);    }    for (int i = 0 ; i < 1 << 4 ; i ++) {        id[i] = i;    }    sort (id , id + (1 << 4) , cmp);}void check(int i) {    if (vis[i]) return ;    flag[i][0] = isprime[i];    int sum = 0 , cnt = 0;    LL product = 1LL;    for (int j = 1 ; j * j <= i ; j ++) {        if(i % j == 0) {            product = (LL)product * j;            sum += j;            cnt ++;            if (j * j != i) {                product *= i / j;                sum += i / j;                cnt ++;            }            if (product == (LL)i * i) {                product = 1LL;            }        }    }    flag[i][1] = isprime[cnt];    flag[i][2] = isprime[sum];    flag[i][3] = issquare(product);    for (int j = 0 ; j < 4 ; j ++) {        score[i] += flag[i][j] ? 1 : 0;        state[i] += (flag[i][j] ? 1 : 0) << j;    }    vis[i] = true;}int main () {    #ifndef ONLINE_JUDGE        freopen("input.txt", "r" , stdin);    #endif    Init();    int t;    scanf ("%d", &t);    while (t --) {        scanf ("%d %d", &n, &k);        for (int i = 0 ; i < 1 << 4; i ++) {            cnt[i] = 0;        }        for (int i = 0 ; i < n ; i ++) {            int a , b;            scanf ("%d %d", &a, &b);            check(a);            cnt[state[a]] += b;            printf("%d%c", score[a], i == n - 1 ? '\n' : ' ');        }        for (int i = 0 ; i < 4 ; i ++) {            scanf ("%d", &b[i]);        }        int ans = - (1 << 20);        for (int i = 0 ; i < 1 << 16 ; i ++) {            int remain = k , ret = 0 , cur = 0;            for (int j = 0 ; j < 1 << 4 ; j ++) {                if(! (i & (1 << id[j]))) continue;                if(cnt[id[j]]) {                    remain --;                    ret += get[id[j]];                }                else remain = -1;            }             if (remain < 0) continue;            for (int j = 0 ; j < 1 << 4 ; j ++) {                if(! (i & (1 << id[j]))) continue;                cur |= id[j];                if(remain && remain > cnt[id[j]] - 1) {                    remain -= cnt[id[j]] - 1;                    ret += (cnt[id[j]] - 1) * get[id[j]];                }                else {                    ret += remain * get[id[j]];                    remain = 0;                }            }            if (remain == 0) {                for (int j = 0 ; j < 4 ; j ++) {                    if(! ((1 << j) & cur) )                        ret += b[j];                }                ans = max (ans , ret);            }        }        printf ("%d\n", ans);    }    return 0;}