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2020-2021:teams:wangzai_milk:20200718比赛记录 [2020/07/22 21:23] infinity37 [题解] |
2020-2021:teams:wangzai_milk:20200718比赛记录 [2020/07/23 20:44] (当前版本) wzx27 |
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===== 题解 ===== | ===== 题解 ===== | ||
- | ==== B - Basic Gcd Problem ==== | + | ==== A - Clam and Fish ==== |
- | 题意:当$x>1$时,$f_c(x)=max_{i=1...x-1}c·f_c(gcd(i,x))$ | + | 依次经过 $n$ 个格子,每个格子上可能有 $\text{fish}$ 或 $\text{clam}$。 |
- | 找到长度 $9$ 和 $6$ 的边,叉集判断顺逆时针关系即可。 | + | 每个格子上可以下面的选一个操作: |
- | <hidden><code cpp> | + | 如果有 $\text{clam}$,可以制作一个鱼饵 |
+ | |||
+ | 如果有 $\text{fish}$,可以得到一条鱼 | ||
+ | |||
+ | 消耗一个鱼饵获得一条鱼 | ||
+ | |||
+ | 什么事都不做 | ||
+ | |||
+ | 显然有鱼的位置选操作三。剩下的有能制作鱼饵就制作鱼饵,否则就用鱼饵捕鱼。最后加上剩下鱼饵的二分之一。 | ||
+ | |||
+ | <hidden code> <code cpp> | ||
#include<bits/stdc++.h> | #include<bits/stdc++.h> | ||
- | #define pii pair<int,int> | + | #define ALL(x) (x).begin(),(x).end() |
#define ll long long | #define ll long long | ||
+ | #define ull unsigned long long | ||
+ | #define pii_ pair<int,int> | ||
+ | #define mp_ make_pair | ||
+ | #define pb push_back | ||
+ | #define fi first | ||
+ | #define se second | ||
+ | #define rep(i,a,b) for(int i=(a);i<=(b);i++) | ||
+ | #define per(i,a,b) for(int i=(a);i>=(b);i--) | ||
+ | #define show1(a) cout<<#a<<" = "<<a<<endl | ||
+ | #define show2(a,b) cout<<#a<<" = "<<a<<"; "<<#b<<" = "<<b<<endl | ||
using namespace std; | using namespace std; | ||
- | #define eps 1e-4 | + | const ll INF = 1LL<<60; |
- | struct Node | + | const int inf = 1<<30; |
+ | const int maxn = 2e6+5; | ||
+ | inline void fastio() {ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);} | ||
+ | |||
+ | char s[maxn]; | ||
+ | int main() | ||
{ | { | ||
- | double x,y; | + | fastio(); |
- | Node(double x=0,double y=0):x(x),y(y){} | + | int _; char x; |
- | Node operator -(Node a){return Node(x-a.x,y-a.y);} | + | for(cin>>_;_;_--){ |
- | }p[22]; | + | int n; cin>>n; |
- | int dcmp(double x) | + | cin>>s+1; |
- | { | + | int ans = 0,cnt = 0; |
- | if(x<-eps)return -1; | + | rep(i,1,n){ |
- | return x>eps; | + | if(s[i]=='0'){ |
+ | if(cnt) ans++,cnt--; | ||
+ | }else if(s[i]=='1'){ | ||
+ | cnt++; | ||
+ | } | ||
+ | else ans++; | ||
+ | } | ||
+ | ans += cnt/2; | ||
+ | cout<<ans<<endl; | ||
+ | } | ||
+ | return 0; | ||
} | } | ||
- | double sqr(double x){return x*x;} | + | |
- | double dissqr(Node a,Node b){return sqr(a.x-b.x)+sqr(a.y-b.y);} | + | </code> </hidden> |
- | double Cross(Node a,Node b){return a.x*b.y-a.y*b.x;} | + | \\ |
+ | |||
+ | ==== B - Classical String Problem ==== | ||
+ | |||
+ | 题意:两种操作,把字符串最后x个接到前面,把字符串最前面x个接到后面。最后询问一次当前字符串。 | ||
+ | |||
+ | 题解:相当于把字符串搞成一个环,有一个指针指着开头,每次操作相当于动指针,就每次加减移动字符数然后模长度就好了。 | ||
+ | |||
+ | <hidden><code cpp> | ||
+ | #include <bits/stdc++.h> | ||
+ | using namespace std; | ||
+ | const int N = 2e6+5; | ||
+ | char s[N]; | ||
int main() | int main() | ||
{ | { | ||
- | int t; | + | int Q; |
- | scanf("%d",&t); | + | scanf("%s",s); |
- | while(t--) | + | scanf("%d",&Q); |
- | { | + | char op[3]; |
- | for(int i=0;i<20;i++)scanf("%lf%lf",&p[i].x,&p[i].y); | + | int n = strlen(s); |
- | int j=0; | + | int st = 0,x; |
- | for(int i=0;i<20;i++) | + | for (int i = 1;i<= Q;i++) { |
- | { | + | scanf("%s%d",op,&x); |
- | if(dcmp(dissqr(p[i],p[(i+1)%20])-81)==0) | + | if (op[0] == 'A') { |
- | {j=i;break;} | + | x--; |
- | } | + | int pos = ((st+x)%n+n)%n; |
- | if(dcmp(dissqr(p[(j+2)%20],p[(j+1)%20])-36)==0) | + | printf("%c\n",s[pos]); |
- | { | + | } else { |
- | if(Cross(p[(j+2)%20]-p[(j+1)%20],p[j]-p[(j+1)%20])<0)puts("right"); | + | st = ((st+x)%n+n)%n; |
- | else puts("left"); | + | } |
- | } | + | } |
- | else | + | |
- | { | + | |
- | if(Cross(p[(j-1+20)%20]-p[j],p[(j+1)%20]-p[j])<0)puts("right"); | + | |
- | else puts("left"); | + | |
- | } | + | |
- | } | + | |
return 0; | return 0; | ||
} | } | ||
行 119: | 行 160: | ||
} | } | ||
} | } | ||
+ | return 0; | ||
+ | } | ||
+ | </code></hidden> | ||
+ | \\ | ||
+ | ==== D - Points Construction Problem ==== | ||
+ | |||
+ | 题意:给一个坐标轴,在整数点上染色,一开始所有的都是白色,要求染n个,使得有m对相邻的点颜色不同。 | ||
+ | |||
+ | 题解:构造,首先可以发现只有m为偶数的时候有方案,然后对于一个新染色点,可能多4对,2对或0对,然后枚举多4对的有几个点,2对的有几个点,算出0对的有几个点,然后先一行空一格染一个,染4对的,然后再最后一个4对的上侧和右侧增加2对的,看剩下的0对的能不能被完全放进这个L形里。可以就可以,不行就没了。 | ||
+ | |||
+ | <hidden><code cpp> | ||
+ | #include <bits/stdc++.h> | ||
+ | using namespace std; | ||
+ | const int N = 55; | ||
+ | int ans[N][2]; | ||
+ | int n,m; | ||
+ | int main() | ||
+ | { | ||
+ | int cas; | ||
+ | scanf("%d",&cas); | ||
+ | while (cas--) { | ||
+ | int n,m; | ||
+ | scanf("%d%d",&n,&m); | ||
+ | int minans = 4*n; | ||
+ | for (int i = 1;i<= n;i++) { | ||
+ | int x=i;int y = n/x; | ||
+ | if (n > x*y) | ||
+ | minans=min(minans,2*(x+y+1)); | ||
+ | else minans = min(minans,2*(x+y)); | ||
+ | } | ||
+ | bool find = false; | ||
+ | for (int cnt4 = 1;cnt4<=n && !find;cnt4++) { | ||
+ | for (int cnt2 = 0;cnt2+cnt4<=n && !find;cnt2++) { | ||
+ | if (cnt2*2+cnt4*4!=m)continue; | ||
+ | int cnt0 = n-cnt4-cnt2; | ||
+ | int x = cnt2/2;int y = cnt2-x; | ||
+ | if (cnt0 <= x*y) { | ||
+ | find = true; | ||
+ | int tmp = 0; | ||
+ | for (int i = 1;i<= cnt4;i++) | ||
+ | ans[++tmp][0] = i<<1; | ||
+ | for (int i = 1;i<= x;i++) { | ||
+ | tmp++; | ||
+ | ans[tmp][0] = ans[cnt4][0]+i; | ||
+ | ans[tmp][1] = ans[cnt4][1]; | ||
+ | } | ||
+ | for (int i = 1;i<= y;i++) { | ||
+ | tmp++; | ||
+ | ans[tmp][0] = ans[cnt4][0]; | ||
+ | ans[tmp][1] = ans[cnt4][1]+i; | ||
+ | } | ||
+ | for (int i = 0;i< cnt0;i++) { | ||
+ | int tx = i%x+1,ty = i/x+1; | ||
+ | tmp++; | ||
+ | ans[tmp][0] = ans[cnt4][0]+tx; | ||
+ | ans[tmp][1] = ans[cnt4][1]+ty; | ||
+ | } | ||
+ | } | ||
+ | } | ||
+ | } | ||
+ | if (find) { | ||
+ | printf("Yes\n"); | ||
+ | for (int i = 1;i<= n;i++) | ||
+ | printf("%d %d\n",ans[i][0],ans[i][1]); | ||
+ | } else { | ||
+ | printf("No\n"); | ||
+ | } | ||
+ | } | ||
return 0; | return 0; | ||
} | } |