UPC第41场，第42场部分题解

第41场

A: LR Constraints

We have N cards arranged in a row from left to right. We will write an integer between 1 and K (inclusive) on each of these cards, which are initially blank.
Given are K restrictions numbered 1 through K. Restriction i is composed of a character ci and an integer ki. If ci is L, the ki-th card from the left in the row must be the leftmost card on which we write
i. If ci is R, the ki-th card from the left in the row must be the rightmost card on which we write i.
Note that for each integer i from 1 through K, there must be at least one card on which we write i.
Find the number of ways to write integers on the cards under the K restrictions, modulo 998244353.

N个卡片，上面的数字为1~k，然后有k个限制条件，每个限制条件有1个字母（ci）和一个数字（ki），如果ci是‘L’ 表示i能够输入的最左边的位置是ki，如果ci是‘R’ 表示i能够输入的最右边的位置是ki。

Constraints
1≤N,K≤1000
ci is L or R.
1≤ki≤N
ki≠kj if i≠j.

输入

Input is given from Standard Input in the following format:
N K
c1 k1
⋮
cK kK

输出

Find the number of ways to write integers on the cards under the K restrictions in the Problem Statement, modulo 998244353.

样例输入

【样例1】
3 2
L 1
R 2
【样例2】
30 10
R 6
R 8
R 7
R 25
L 26
L 13
R 14
L 11
L 23
R 30

样例输出

【样例1】
1
【样例2】
343921442

按题目要求模拟,之后用乘法原理累计贡献即可。

code:code:code:

#include<iostream>
#include<iomanip>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<algorithm>
#define ll long long
using namespace std;
const int maxn=10005,mod = 998244353;
int vis1[maxn],vis2[maxn];
ll n,k,k1;
ll s[maxn];
char ch;
int main()
{cin >> n >> k;for(int i=1;i<=k;i++){cin >> ch >> k1;vis1[k1]=1;if(ch=='L'){for(ll i=k1+1;i<=n;i++){s[i]++;}}else{for(ll i=1;i<k1;i++){s[i]++;}}}ll ans=1;for(ll i=1;i<=n;i++){if(!vis1[i]) ans=(ans*s[i])%mod;}cout << ans;return 0;
}

B: LCM of GCDs

We have a red bag, a blue bag, and N card packs. Initially, both bags are empty. Each card pack contains two cards with integers written on them. We know that the i-th card pack contains a card with ai and another with bi.
For each card pack, we will put one of its cards in the red bag, and the other in the blue bag.
After we put all cards in the bags, let X be the greatest common divisor of all integers written on cards in the red bag. Similarly, let Y be the greatest common divisor of all integers written on cards in the blue bag. Our score will be the least common multiple of X and Y.

Find the maximum possible score.

我们有一个红色的包，一个蓝色的包和N个卡片包。最初，两个袋子都是空的。每个卡片包包含两张写有整数的卡片。我们知道，第i个卡包包含一张ai卡和另一张bi卡。

对于每个卡片包，我们将把它的一张卡片放在红色的袋子里，另一张放在蓝色的袋子里。

当我们把所有的卡片放在袋子里之后，让X成为红色袋子里卡片上所有整数的最大公约数。类似地，让Y是写在蓝包卡片上的所有整数的最大公约数。我们的分数将是X和Y的最小公倍数。

找出可能的最大分数。

Constraints
All values in input are integers.
1≤N≤50
1≤ai,bi≤109

输入

Input is given from Standard Input in the following format:
N
a1 b1
⋮
aN bN

输出

Print the maximum possible score.

样例输入

【样例1】
2
2 15
10 6
【样例2】
5
148834018 644854700
947642099 255192490
35137537 134714230
944287156 528403260
68656286 200621680
【样例3】
20
557057460 31783488
843507940 794587200
640711140 620259584
1901220 499867584
190122000 41414848
349507610 620259584
890404700 609665088
392918800 211889920
507308870 722352000
156850650 498904448
806117280 862969856
193607570 992030080
660673950 422816704
622015810 563434560
207866720 316871744
63057130 117502592
482593010 366954816
605221700 705015552
702500790 900532160
171743540 353470912

样例输出

【样例1】
10
【样例2】
238630
【样例3】
152594452160

注意到N<=50，每次的选择只有两种情况，a[i],b[i]a[i],b[i]a[i],b[i],or b[i],a[i]b[i],a[i]b[i],a[i],不妨直接暴力dfs，这里为了阻止搜索过程中的无效回溯，使用集合标记我们做过的过程，可以大大减少时间。

codecodecode:

#include<iostream>
#include<iomanip>
#include<cstring>
#include<cmath>
#include<cstdio>
#include <vector>
#include <set>
#include<algorithm>
#define ll long long
using namespace std;
const int maxn=105,mod = 998244353;
int a[maxn], b[maxn];
int n;
ll ans = 1;
set<pair<pair<int,int>,int>>st;
void dfs(ll d1,ll d2,int c)
{if(st.count({{d1, d2}, c})){return ;}if(c==n+1){ans=max(ans,d1/__gcd(d1,d2)*d2);return;}st.insert({{d1,d2},c});dfs(__gcd(d1,(ll)a[c]), __gcd(d2,(ll)b[c]), c+1);dfs(__gcd(d1,(ll)b[c]), __gcd(d2,(ll)a[c]), c+1);
}
int main()
{cin >> n;for(int i=1;i<=n;i++){scanf("%d%d", &a[i], &b[i]);}dfs(a[1],b[1],2);cout << ans ;
}

D.Garbage Classification

On the CVBB planet, garbage classification has been gradually executed to help save resources and protect the environment. Nowadays people have to be equipped with knowledge of distinguishing different types of garbage. Now, given the waste compositions of a discarded product, you are asked to determine which category it belongs to.

The waste compositions are represented as a string s consisting of only lowercase letters, where each letter represents a waste composition and has an equal proportion. Each waste composition in that product is in one of the three situations, dry, wet or harmful.

The product can be classified by the following rules:

In case that at least 25% of its compositions is harmful, it is harmful garbage.
In case that at most 10% of its compositions is harmful, it is recyclable garbage.
In other cases, if the proportion of dry compositions in it is at least twice that of wet compositions, it is dry garbage, or otherwise, it is wet garbage.

输入

There are multiple test cases. The first line contains an integer T (1≤T≤50), indicating the number of test cases. Test cases are given in the following.

For each test case, the first line contains a non-empty string s (1≤∣s∣≤2000) consisting of only lowercase letters.

The second line contains a string t of length 26, consisting of only characters in {d,w,h}. The i-th character of t represents the situation of the waste composition denoted by the i-th lowercase letter, where ‘d’, ‘w’ and ‘h’ mean dry, wet or harmful respectively.

输出

For each test case, output “Case #x: y” in one line (without quotes), where x indicates the case number starting from 1, and y (y∈{Harmful,Recyclable,Dry,Wet}) denotes the garbage type of the product in this test case.

样例输入

4
dqxxjefgctjgdbqxphff
hddhddhhhdhdhwwddhhdwwdhhw
iqdvfzzdqsbdevzebego
wdhddwwhwhdhhhwwdwdhwdwhhd
erkjqzsmchcmbqeasadf
dwddddwdwdwhdwhhdhhwwhhwdh
mcxkwmxxlhbrymwawhio
ddhwhddhwwwdddwdwhwwwhdwdw

样例输出

Case #1: Harmful
Case #2: Recyclable
Case #3: Dry
Case #4: Wet

还是模拟题。。。

code:code:code:

#include<iostream>
#include<iomanip>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<algorithm>
#define ll long long
using namespace std;
const int maxn=100005;
int n,cnt=0;
int m;
int main()
{string s,s1;cin >> m;while(m--){cin >> s1;cin >> s;double sum1=0,sum2=0,sum3=0;int n=s1.length();for(int i=0;i<n;i++){int k=s1[i]-'a';if(s[k]=='d'){sum1++;}else if(s[k]=='w'){sum2++;}else if(s[k]=='h'){sum3++;}}printf("Case #%d: ",++cnt);//cout << sum1 << " " << sum2 << " " << sum3 << endl;if(sum3/n>=0.25){printf("Harmful\n");}else if(sum3/n<=0.1){printf("Recyclable\n");}else{if(sum1>=sum2*2){printf("Dry\n");}else{printf("Wet\n");}}}return 0;
}

L.铺设道路

春春是一名道路工程师，负责铺设一条长度为n的道路。
铺设道路的主要工作是填平下陷的地表。整段道路可以看作是n块首尾相连的区域，一开始，第i块区域下陷的深度为di。
春春每天可以选择一段连续区间[L,R]，填充这段区间中的每块区域，让其下陷深度减少1。在选择区间时，需要保证，区间内的每块区域在填充前下陷深度均不为0。
春春希望你能帮他设计一种方案，可以在最短的时间内将整段道路的下陷深度都变为0。

输入

输入包含两行，第一行包含一个整数n，表示道路的长度。
第二行包含n个整数，相邻两数间用一个空格隔开，第i个整数为di。

输出

输出仅包含一个整数，即最少需要多少天才能完成任务。

样例输入

6
4 3 2 5 3 5

样例输出

贪心，如果相邻后者小于前者，前者的填充一定会覆盖后者；反之，后者就需要再进行铺设，铺设的量为a[i]-a[i-1]。

code:code:code:

#include<iostream>
#include<iomanip>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<algorithm>
#define ll long long
using namespace std;
const int maxn=100005;
int n,tot=0;
ll a[maxn];
int main()
{int n;ll ans=0;cin >> n;for(int i=1;i<=n;i++){cin >> a[i];}for(int i=1;i<=n+1;i++){if(i!=1){if(a[i]<a[i-1]){ans+=abs(a[i]-a[i-1]);}}}cout << ans << endl;return 0;
}

F: Is Today Friday?

No, it’s not Friday