[Aizu] ITP2_10_C~D: bit flag and bit mask

前言

ITP系列之位运算, 具体内容参见bitset

题目链接

ITP2_10_C: Bit Flag
ITP2_10_D: Bit Mask

求解

第一题

明确的指定了让处理哪一位, 很容易就用bitset类搞定了

#include <bits/stdc++.h>
using namespace std;
 
int main(void) {
    ios::sync_with_stdio(false);
    cin.tie(0);
 
    int q, com, i;
    bitset<64> bs;
    bs.reset();
 
    cin >> q;
    while (q--) {
        cin >> com;
        switch (com) {
            case 0:
                cin >> i;
                cout << bs.test(i) << endl;
                break;
            case 1:
                cin >> i;
                bs.set(i);
                break;
            case 2:
                cin >> i;
                bs.reset(i);
                break;
            case 3:
                cin >> i;
                bs.flip(i);
                break;
            case 4:
                cout << (bs.all() ? 1 : 0) << endl;
                break;
            case 5:
                cout << (bs.any() ? 1 : 0) << endl;
                break;
            case 6:
                cout << (bs.none() ? 1 : 0) << endl;
                break;
            case 7:
                cout << bs.count() << endl;
                break;
            case 8:
                cout << bs.to_ulong() << endl;
        }
    }
}

第二题

题目中说明了, 不再是给定一位进行处理, 而是给定一部分(掩码), 然后每次都要对这一部分进行对应的处理, 我第一次尝试的时候是通过创建一个嵌套的vector容器来存储那些特殊值的, 在看了别人的代码后, 才尝试着去使用bitset作为数组元素来用.

第一次代码

#include <bits/stdc++.h>
using namespace std;
 
int main(void) {
    ios::sync_with_stdio(false);
    cin.tie(0);
 
    vector<vector<int> > vec;
    vector<int> ve;
    vector<int>::iterator it;
    bitset<64> bs;
    int n, q, com, i, k;
    int flag, cnt;
    unsigned long val;
 
    cin >> n;
    while (n--) {
        ve.clear();
        cin >> k;
        while (k--) {
            cin >> i;
            ve.push_back(i);
        }
        vec.push_back(ve);
    }
 
    cin >> q;
    while (q--) {
        cin >> com >> i;
        switch (com) {
            case 0:
                cout << bs.test(i) << endl;
                break;
            case 1:
                ve = vec[i];
                for (it = ve.begin(); it != ve.end(); it++) {
                    bs.set(*it);
                }
                break;
            case 2:
                ve = vec[i];
                for (it = ve.begin(); it != ve.end(); it++) {
                    bs.reset(*it);
                }
                break;
            case 3:
                ve = vec[i];
                for (it = ve.begin(); it != ve.end(); it++) {
                    bs.flip(*it);
                }
                break;
            case 4:
                ve = vec[i];
                flag = 1;
                for (it = ve.begin(); flag && it != ve.end(); it++) {
                    flag = bs.test(*it);
                }
                cout << flag << endl;
                break;
            case 5:
                ve = vec[i];
                flag = 0;
                for (it = ve.begin(); flag == 0 && it != ve.end(); it++) {
                    flag = bs.test(*it);
                }
                cout << flag << endl;
                break;
            case 6:
                ve = vec[i];
                flag = 0;
                for (it = ve.begin(); flag == 0 && it != ve.end(); it++) {
                    flag = bs.test(*it);
                }
                if (flag) cout << 0 << endl;
                else cout << 1 << endl;
                break;
            case 7:
                ve = vec[i];
                cnt = 0;
                for (it = ve.begin(); it != ve.end(); it++) {
                    cnt += bs.test(*it);
                }
                cout << cnt << endl;
                break;
            case 8:
                ve = vec[i];
                val = 0;
                for (it = ve.begin(); it != ve.end(); it++) {
                    val += ((unsigned long)bs.test(*it)) << *it;
                }
                cout << val << endl;
        }
    }
}

第二次学习别人风格的代码

#include <bits/stdc++.h>
using namespace std;
 
int main(void) {
    ios::sync_with_stdio(false);
    cin.tie(0);
 
    vector<bitset<64> > v(10);
    bitset<64> bs;
 
    int n, q, k, com, i;
 
    cin >> n;
    for (int j = 0; j < n; j++) {
        cin >> k;
        while (k--) {
            cin >> i;
            v[j][i] = 1;
        }
    }
 
    cin >> q;
    while (q--) {
        cin >> com >> i;
        if (com == 0) cout << bs.test(i) << endl;
        if (com == 1) bs |= v[i];
        if (com == 2) bs &= ~v[i];
        if (com == 3) bs ^= v[i];
        if (com == 4) cout << ((bs & v[i]) == v[i]) << endl;
        if (com == 5) cout << ((bs & v[i]).to_ulong() != 0) << endl;
        if (com == 6) cout << ((bs & v[i]).to_ulong() == 0) << endl;
        if (com == 7) cout << (bs & v[i]).count() << endl;
        if (com == 8) cout << (bs & v[i]).to_ulong() << endl;
    }
}