Codeforces #320（div2）

最新推荐文章于 2022-05-09 10:55:44 发布

TaoSama

最新推荐文章于 2022-05-09 10:55:44 发布

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分类专栏：搜索 - 二/三分搜索

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本文解析了五道经典编程题目，包括细菌繁殖、团队组建、折线问题、或运算游戏及弱点与贫弱值等，提供了详细的算法思路与实现代码。

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A. Raising Bacteria

题意: 每次放1 每天翻1倍问放几个恰好达到n

分析: 考虑快速幂 - - 然后就可以了其实直接__builtin_popcount(x)就是答案

代码:

//
//  Created by TaoSama on 2015-09-16
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

int n, ans;

void dfs(int x) {
    if(!x) return;
    if(x & 1) ++ans;
    dfs(x >>= 1);
}

int main() {
#ifdef LOCAL
    freopen("in.txt", "r", stdin);
//  freopen("out.txt","w",stdout);
#endif
    ios_base::sync_with_stdio(0);

    while(scanf("%d", &n) == 1) {
        ans = 0;
        dfs(n);
        printf("%d\n", ans);
    }
    return 0;
}

B. Finding Team Member
题意: 组成n队然后按照题目信息给了一些组队信息然后每次组队都得组最大的

分析: 排个序标记下组过没 - - 然后记录答案

代码:

//
//  Created by TaoSama on 2015-09-16
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

int n, ans[805];
struct P {
    int s, i, j;
    bool operator<(const P& rhs) const {
        return s > rhs.s;
    }
} a[800 * 800];

int main() {
#ifdef LOCAL
    freopen("in.txt", "r", stdin);
//  freopen("out.txt","w",stdout);
#endif
    ios_base::sync_with_stdio(0);

    while(scanf("%d", &n) == 1) {
        int cnt = 0;
        n <<= 1;
        for(int i = 2; i <= n; ++i) {
            for(int j = 1; j < i; ++j) {
                scanf("%d", &a[cnt].s);
                a[cnt].i = i;
                a[cnt].j = j;
//                printf("%d %d %d\n", a[cnt].s, a[cnt].i, a[cnt].j);
                ++cnt;
            }
        }
        sort(a, a + cnt);
        memset(ans, 0, sizeof ans);
        for(int i = 0; i < cnt; ++i) {
            P& cur = a[i];
//          printf("%d %d %d\n", cur.s, cur.i, cur.j);
            if(ans[cur.i] || ans[cur.j]) continue;

            ans[cur.i] = cur.j;
            ans[cur.j] = cur.i;
        }
        for(int i = 1; i <= n; ++i)
            printf("%d%c", ans[i], " \n"[i == n]);
    }
    return 0;
}

C. A Problem about Polyline

题意: /\/\/\/\ 这种形状然后给一个(x,y)坐标求这个坐标在上面的最小的高

分析: 斜率是1 所以所在的端点是(x-y,0) 和 (x+y,0) 假设这个点在峰顶

左边有c1=(x-y)/(2y)个周期右边有c2=(x+y)/(2y)个周期

然后(x-y) / c1 / 2就是答案 - -

x < y是显然没答案的 - - 特判下x==y 不然会输出除0输出nan

代码:

//
//  Created by TaoSama on 2015-09-16
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;
const double EPS = 1e-10;

int x, y;

int main() {
#ifdef LOCAL
    freopen("in.txt", "r", stdin);
//  freopen("out.txt","w",stdout);
#endif
    ios_base::sync_with_stdio(0);

    while(scanf("%d%d", &x, &y) == 2) {
        if(x < y) puts("-1");
        else if(x == y) printf("%.12f\n", 1.0 * x);
        else {
            double k1 = x - y, k2 = x + y;
            int c1 = k1 / 2 / y, c2 = k2 / 2 / y;
            double ans = min(k1 / 2 / c1, k2 / 2 / c2);
            printf("%.12f\n", ans);
        }
    }
    return 0;
}

D. "Or" Game

题意: 给定一个序列a1,a2,...an 现在可以最多乘k次x 之后求整个序列的最大xor值

分析: 显然肯定乘到一个上面最优但是要枚举乘的数不能乘最大的有反例

比如:

3 1 2

4 5 2

要乘到4上面而不是5

代码:

//
//  Created by TaoSama on 2015-09-17
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 2e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

int n, k, x;
long long a[N], l[N], r[N];

int main() {
#ifdef LOCAL
    freopen("in.txt", "r", stdin);
//  freopen("out.txt","w",stdout);
#endif
    ios_base::sync_with_stdio(0);

    while(scanf("%d%d%d", &n, &k, &x) == 3) {
        l[0] = r[n + 1] = 0;
        for(int i = 1; i <= n; ++i) {
            scanf("%I64d", a + i);
            l[i] = l[i - 1] | a[i];
        }
        for(int i = n; i; --i)
            r[i] = r[i + 1] | a[i];

        long long ans = 0;
        for(int i = 1; i <= n; ++i) {
            long long cur = l[i - 1] | r[i + 1], tmp = a[i];
            for(int j = 1; j <= k; ++j) tmp *= x;
            ans = max(ans, cur | tmp);
        }
        printf("%I64d\n", ans);
    }
    return 0;
}

E. Weakness and Poorness

题意: 求一个东西然后自己看题吧 - - 反正就四行

分析: 赛上健忘症又犯了读题忘一开始check里面写的max 然后读了output发现是求min然后就改成了min

也不想想刚才自己过了样例啊 - - 肯定是没写错的这种时候应该通读一遍题啊一直以为三分写错了简直逗比不然是可以过的啊

绝对值函数是凸函数然后凸函数的max也是凸函数 (详情见百度百科这里我是一直记的结论)

好吧很显然应该是三分然后让你求的东西明显是最大子串和 dp就好了因为是绝对值把负数的也搞一遍

三分范围别忘记了负的

代码:

//
//  Created by TaoSama on 2015-09-17
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 2e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;
const double eps = 1e-8;

int n;
double a[N], b[N];

double check(double x) {
    for(int i = 1; i <= n; ++i) b[i] = a[i] - x;
    double sum = 0, ans1 = 0, ans2 = 0;
    for(int i = 1; i <= n; ++i) {
        sum += b[i];
        if(sum > ans1) ans1 = sum;
        if(sum < 1e-10) sum = 0;
    }
    sum = 0;
    for(int i = 1; i <= n; ++i) {
        sum += b[i];
        if(sum < ans2) ans2 = sum;
        if(sum > 1e-10) sum = 0;
    }
    return max(ans1, -ans2);
}

int main() {
#ifdef LOCAL
    freopen("in.txt", "r", stdin);
//  freopen("out.txt","w",stdout);
#endif
    ios_base::sync_with_stdio(0);

    while(scanf("%d", &n) == 1) {
        for(int i = 1; i <= n; ++i) scanf("%lf", a + i);

        double l = -1e10, r = 1e10;
        for(int i = 1; i <= 300; ++i) {
            double ll = (l + r) / 2, rr = (ll + r) / 2;
            if(check(ll) < check(rr)) r = rr;
            else l = ll;
        }
//        printf("%.10f\n", l);
        printf("%.10f\n", check(l));
    }
    return 0;
}