2018-2019 ACM-ICPC Nordic Collegiate Programming Contest (NCPC 2018) - 4.28

 

赛后补了几道

赛中我就写了两个...

A - Altruistic AmphibiansGym - 101933A

看了眼榜没几个人做。就没看。

最后发现就是一个DP(但是我觉得复杂度有点迷)

题意:$n$只青蛙有参数$l,w,h$分别表示弹跳力,体重,身高,在一口深为$d$的井里

一只青蛙不能承受比他重的重量,问最多有多少只能出去(达到高度严格大于d)

重的肯定比轻的晚出去,那么轻的肯定由重的来转移,所以先排序,从重到轻的排

$dp_{i}$表示体重为i最高能叠多高 瞎jb转移一下就好了

#include 
#include 
#include 
#define ll long long
using namespace std;

inline int read() {
    int x = 0, f = 1; char ch = getchar();
    while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); }
    while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
    return x * f;
}

const int N = 1e5 + 10;
const int M = 1e8 + 10;
struct P {
    int l, w, h;
    bool operator < (const P &rhs) const {
        return w > rhs.w;
    }
} p[N];
int dp[M], n, d, ans;

int main() {
    n = read(), d = read();
    for (int i = 1; i <= n; i++) p[i].l = read(), p[i].w = read(), p[i].h = read();
    ans = 0;
    sort(p + 1, p + n + 1);
    for (int i = 1; i <= n; i++) {
        if (dp[p[i].w] + p[i].l > d) ans++;
        for (int j = p[i].w + 1; j < min(p[i].w * 2, M); j++) {
            dp[j - p[i].w] = max(dp[j-p[i].w], dp[j] + p[i].h);
        } 
    }
    printf("%d\n", ans);
    return 0;
}
View Code

 

B - Baby Bites Gym - 101933B

#include 
#include 
using namespace std;

const int N = 1010;
int a[N];
int n;
char s[N];

int main() {
    scanf("%d", &n);
    bool ans = true;
    for (int i = 1; i <= n; i++) {
        scanf("%s", s);
        int len = strlen(s);
        if (s[0] == 'm') a[i] = i;
        else {
            int x = 0;
            for (int j = 0; j < len; j++) x = x * 10 + s[j] - '0';
            a[i] = x;
            if (x != i) {
                ans = false;
            }
        }
    }
    puts(ans?"makes sense":"something is fishy");
    return 0;
}
View Code

 

C - Code Cleanups Gym - 101933C

阅读理解题啊。自己瞎糊了一发。读不下去。就丢给队友了。

不管了。

 

E - Explosion Exploit Gym - 101933E

题意:分别有$n,m$个士兵,每个士兵有一个血量,有d个攻击,等概率分给每一个士兵。

问敌方士兵全死(m)的概率是多少

队友过的。学习了新知识,概率记忆化+状压

用一个long long来表示状态

unordered_map来存状态对应的概率 再回溯搜索即可 tql

#include 
#define ll long long
using namespace std;

inline int read() {
    int x = 0, f = 1; char ch = getchar();
    while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); }
    while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
    return x * f;
}

unordered_mapdouble> mp;
int a[2][10];

ll getsta() {
    ll ret = 0;
    for (int i = 1; i <= 6; i++) ret = ret * 10 + 1LL * a[1][i];
    for (int i = 1; i <= 6; i++) ret = ret * 10 + 1LL * a[0][i];
    return ret;
}

double dfs(ll sta, int d) {
    if (mp.count(sta)) return mp[sta];
    if (sta < 1000000) return 1;
    if (d == 0) return 0;
    int sum = 0;
    for (int i = 0; i < 2; i++) 
        for (int j = 1; j <= 6; j++)
            sum += a[i][j];
    double ret = 0;
    for (int i = 0; i < 2; i++) {
        for (int j = 1; j <= 6; j++) {
            if (!a[i][j]) continue;
            a[i][j]--;
            a[i][j-1]++;
            ll s = getsta();
            double res = dfs(s, d - 1);
            a[i][j]++;
            a[i][j-1]--;
            mp[s] = res;
            ret += a[i][j] * 1.0 / sum * res;
        }
    }
    return ret;
}

int main() {
    int n = read(), m = read(), d = read();
    for (int i = 1, x; i <= n; i++) x = read(), a[0][x]++;
    for (int i = 1, x; i <= m; i++) x = read(), a[1][x]++;
    double res = dfs(getsta(), d);
    printf("%.8f\n", res);
    return 0;
}
View Code

 

H - House Lawn Gym - 101933H

题意:有m台机器,每台机器有名字,价格p,每分钟能工作多少c,充一次电能工作多久t,充电需要多久r

有l面积的地待作,问平均每周能工作一次的机器中价格最小的那个,相同的按输入顺序输出

队友把10080说成10800能忍?

平均一下直接就把充电需要的时间给平均进来 得到每分钟工作多少的p'

再用$l/p'$和10080比就得出答案了

可能难就难在输入部分吧。

#include 
using namespace std;

struct Node
{
    string name;
    int p , c, t, r;
    double cc;
}b[105];
bool vis[105];

int main()
{
    ios::sync_with_stdio(false);
    int m;
    int l;
    cin >> l >> m;
    string a;
    getline(cin,a);
    for(int i=1;i<=m;i++)
    {
        getline(cin,a);
        int sta = 0;
        b[i].name = "";
        b[i].p = b[i].c = b[i].r = b[i].t = 0;
        for(int j=0;j)
        {
            if(a[j]==',')
            {
                sta++;
                continue;
            }
            if(sta == 0)
            {
                b[i].name+=a[j];
            }
            if(sta == 1)
            {
                b[i].p*=10;
                b[i].p+=a[j]-'0';    
            }
            if(sta == 2)
            {
                b[i].c*=10;
                b[i].c+=a[j]-'0';    
            }
            if(sta == 3)
            {
                b[i].t*=10;
                b[i].t+=a[j]-'0';    
            }
            if(sta == 4)
            {
                b[i].r*=10;
                b[i].r+=a[j]-'0';    
            }
        }
    }
    int ans = 1e9;
    for (int i = 1; i <= m; i++) {
        b[i].cc = (b[i].c * b[i].t) * 1.0 / (b[i].t + b[i].r);
        if (l / b[i].cc <= 10080) {
            ans = min(ans, b[i].p);
            vis[i] = 1;
        }
    }
    if (ans == (int)1e9) puts("no such mower");
    else {
        for (int i = 1; i <= m; i++) {
            if (vis[i] && ans == b[i].p) {
                cout << b[i].name << '\n';
            }
        }    
    }
    return 0;
}
View Code

 

J - Jumbled String Gym - 101933J

题意: 0 1串 给你00出现的次数a 01出现的次数b 10出现的次数c 11出现的次数d

问能否构造出01串

WA了好几发 一度崩溃

首先由a d能推出0和1的个数 必定是一个C(n, 2) 把a和d乘二开根 和加一相乘是否等于2a和2d来判断

第二部分

用两个数组表示

$a_{i}$表示第i个0后面出现了几个1

$b_{i}$表示第i个0前面出现了几个1

必定有$a_{i} + b_{i} = cnt_{1}$        $a_{i}\geq a_{i+1}$        $b_{i}\leq b_{i+1}$

$\sum ^{cnt_{0}}_{i=1}a_{i} = b$            $\sum ^{cnt_{0}}_{i=1}b_{i} = c$

所以$b + c = cnt_{0}\times cnt_{1}$才有解

随便举几个例子发现贪心的构造均能满足答案

如样例 3 4 2 1

得到$cnt_{0} = 3$   $cnt_{1} = 2$

$a_{i}$ : 2 2 0

$b_{i}$ : 0 0 2

得到 00110 也符合

所以直接构造就完了

不过要注意

0 0 0 0直接输出0或1就可以了

a为0或d为0的情况 如果$b + c = 0$ 那么对应的0的个数或1的个数为0 否则才为1

然后瞎jb输出就完了

#include 
#include 
#include 
using namespace std;

inline int read() {
    int x = 0, f = 1; char ch = getchar();
    while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar();}
    while (ch >= '0' && ch <= '9') { x = x * 10 + ch - 48; ch = getchar(); }
    return x * f;
}

int main() {
    int a = read(), b = read(), c = read(), d = read();
    int sqra = sqrt(2 * a), sqrd = sqrt(2 * d);
    bool ans = true;
    if ((a + b + c + d) == 0) {
        puts("0");
        return 0;
    }
    if (sqra * (sqra + 1) != a * 2 || sqrd * (sqrd + 1) != d * 2) {
        ans = false;
    } else {
        int cnt0 = sqra, cnt1 = sqrd;
        cnt0++, cnt1++;
        if (cnt1 == 1 && (b + c) == 0) cnt1 = 0;
        if (cnt0 == 1 && (b + c) == 0) cnt0 = 0;
        //printf("%d %d\n", cnt0, cnt1);
        if (b + c != cnt0 * cnt1) {
            ans = false;
        } else {
            if (cnt1 == 0) {
                while (cnt0--) putchar('0');
                return 0;
            }
            if (cnt0 == 0) {
                while (cnt1--) putchar('1');
                return 0;
            }
            int k = b / cnt1;
            for (int i = cnt0; i > cnt0 - k; i--) putchar('0');
            cnt0 -= k;
            k = b - k * cnt1;
            if (k) {
                k = cnt1 - k;
                for (int i = cnt1; i > cnt1 - k; i--) putchar('1');
                cnt1 -= k;
                putchar('0'), cnt0--;
            }    
            while (cnt1--) putchar('1');
            while (cnt0--) putchar('0');
            return 0;
        }
    }
    if (!ans) puts("impossible");
    return 0; 
}
View Code

 

K - King's Colors Gym - 101933K

题意:一棵树n个节点,k种颜色,问有多少种方案用上k个颜色并且相邻两节点颜色不同

我以为要用树形dp做,赛后看题解才明白是个容斥。

用k种的情况是$k\times \left( k-1\right) ^{n-1}$然后其中包含了只用了k-1种 只用了k-2种...的情况

容斥一下就好了

#include 
#define ll long long
using namespace std;

inline ll read() {
    ll x = 0, f = 1; char ch = getchar();
    while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); }
    while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
    return x * f;
}

const ll mod = 1e9 + 7;
const int N = 2550;
ll C[N][N];
void init() {
    for (int i = 0; i < N; i++) C[i][0] = 1, C[i][1] = i;
    for (int i = 1; i < N; i++)
        for (int j = 2; j <= i; j++)
            C[i][j] = (C[i-1][j-1] + C[i-1][j]) % mod;
}

ll qm(ll a, ll b) {
    ll res = 1;
    while (b) {
        if (b & 1) res = res * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return res;
}

int main() {
    init();
    ll n = read(), k = read();
    for (int i = 1; i < n; i++) read();
    ll ans = 0, flag = 1;
    for (int i = k; i >= 1; i--) {
        ll temp = flag * i * qm((ll)i - 1, n - 1) % mod * C[k][i];
        ans = (ans + temp + mod) % mod;
        flag = -flag;
    }        
    printf("%lld\n", ans);
    return 0;
}
View Code

 

转载于:https://www.cnblogs.com/Mrzdtz220/p/10788014.html

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