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

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 <cstdio>
#include <algorithm>
#include <cstring>
#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;
}

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B - Baby Bites Gym - 101933B

#include <cstdio>
#include <cstring>
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;
}

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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 <bits/stdc++.h>
#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_map<ll, double> 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;
}

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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 <bits/stdc++.h>
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<a.length();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;
}

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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 <cstdio>
#include <algorithm>
#include <cmath>
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;
}

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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 <bits/stdc++.h>
#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;
}

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