2021浙江省赛题解（A,C,F,G,J,L,M）

A.League of Legends

题解

签到题

直接求和判断一下

注意会爆 i n t int int以及相等的情况。

代码

#include <bits/stdc++.h>
#define PI atan(1.0)*4
#define rp(i,s,t) for (register int i = (s); i <= (t); i++)
#define RP(i,t,s) for (register int i = (t); i >= (s); i--)
#define sc(x) scanf("%d",&x)
#define scl(x) scanf("%lld",&x)
#define ll long long
#define ull unsigned long long
#define mst(a,b) memset(a,b,sizeof(a))
#define lson rt<<1,l,m
#define rson rt<<1|1,m+1,r
#define pii pair<int,int>
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define m_p make_pair
#define p_b push_back
#define ins insert
#define era erase
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f3f3f3f
#define dg if(debug)
#define pY puts("YES")
#define pN puts("NO")
#define outval(a) cout << "Debuging...|" << #a << ": " << a << "\n";
#define outval2(a,b) cout << "Debuging...|" << #a << ": " << a <<"\t"<< #b << ": " << b << "\n";
#define outval3(a,b,c) cout << "Debuging...|" << #a << ": " << a <<"\t"<< #b << ": " << b <<"\t"<< #c << ": " << c << "\n";
using namespace std;
int debug = 0;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;
}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;
}
inline int read(){int s=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-') f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){s=s*10+ch-'0';ch=getchar();}return s*f;
}
const int N = 2e5+7;
int a[10],b[10];
void solve(){rp(i,1,5) a[i]=read();rp(i,1,5) b[i]=read();ll sum1=0,sum2=0; rp(i,1,5) sum1+=a[i],sum2+=b[i];if(sum1<sum2) cout<<"Red"<<endl;else cout<<"Blue"<<endl;
}
int main(){//ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#ifdef ONLINE_JUDGE
#elsefreopen("in.txt", "r", stdin);//debug = 1;
#endif//time_t beg, end;//if(debug) beg = clock();solve();/*if(debug) {end = clock();printf("time:%.2fs\n", 1.0 * (end - beg) / CLOCKS_PER_SEC);}*/return 0;
}

C.Cube

题解

刚开始以为题意是判断是否是长方体。

然后写了长方体的判断方法，后面看样例发现是正方体后想到一个简单做法。

判断是否有 12 12 12个相等的棱， 12 12 12个相等的面对角线， 4 4 4个相等的体对角线，以及棱，面对角线和体对角线可以形成直角三角形（满足勾股定理）就行了。

代码

#include <bits/stdc++.h>
#define _for(i, a) for(int i = 0, le = (a); i < le; ++i)
#define _rep(i, a, b) for(int i = (a), le = (b); i <= le; ++i)
typedef long long LL;
#define INF 0x3f3f3f3f3f3f3f3f
using namespace std;LL read() {LL x(0), f(1); char c = getchar();while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}while(c >= '0' && c <= '9') {x = x * 10 + c - '0';c = getchar();}return x * f;
}struct poi {int x, y, z;poi(){}poi(int x, int y, int z):x(x), y(y), z(z) {}void sc() {x = read(), y = read(), z = read();}bool operator<(const poi b) const {if(x != b.x) return x < b.x;if(y != b.y) return y < b.y;if(z != b.z) return z < b.z;return 0;}
};int T;
vector<poi> a;
set<poi> a4, ano4;
LL MaxDis;int isEqu(LL a, LL b) {return a == b;
}LL getDis(poi a, poi b) {return (a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y) + (a.z-b.z)*(a.z-b.z);
}int isSqu(poi a, poi b, poi c, poi d) {vector<poi> t;t.push_back(a);t.push_back(b);t.push_back(c);t.push_back(d);vector<LL> vals;_for(i, 4) {_for(j, i) {vals.push_back(getDis(t[i], t[j]));}}sort(vals.begin(), vals.end(), greater<LL>());if(isEqu(vals[0], vals[2])) return 0;if(isEqu(vals[0], MaxDis)) return 0;if(!isEqu(vals[0], vals[1])) return 0;if(!isEqu(vals[2], vals[3])) return 0;if(!isEqu(vals[4], vals[5])) return 0;if(!isEqu(vals[4], vals[2])) return 0;return 1;
}set<poi> getSqu() {set<poi> ans;_for(i, 8) {_for(j, i) {_for(k, j) {_for(l, k) {if(isSqu(a[i], a[j], a[k], a[l])) {ans.insert(a[i]);ans.insert(a[j]);ans.insert(a[k]);ans.insert(a[l]);return ans;}}}}}return ans;
}LL getMinDis(poi a) {LL ans = INF;for(poi i : ano4) {ans = min(getDis(a, i), ans);}return ans;
}void init() {MaxDis = -1;a4.clear();ano4.clear();
}int sol() {init();_for(i, 8) {_for(j, i) {if(!(a[i] < a[j]) && !(a[j] < a[i])) return 0;}}vector<LL> diss;_for(i, 8) {_for(j, i) {diss.push_back(getDis(a[i], a[j]));}}sort(diss.begin(), diss.end());_for(i, 12) {_for(j, i) {if(diss[i] != diss[j]) return 0;}}_for(i, 12) {_for(j, i) {if(diss[12 + i] != diss[12 + j]) return 0;}}_for(i, 4) {_for(j, i) {if(diss[24 + i] != diss[24 + j]) return 0;}}if(diss[0] + diss[12] != diss[24]) return 0;return 1;
}int main() {// freopen("in.txt", "r", stdin);scanf("%d", &T);_for(i, T) {a.resize(8);_for(i, 8) a[i].sc();printf("%s\n", sol() ? "YES":"NO");}return 0;
}

F.Fair Distribution

题解

不难发现对于 n > m n>m n>m的情况直接输出 n − m n-m n−m即可。
对于其他的情况我们可以推出一个公式。

推导过程如下：
我们假设 n n n减小成了 i i i（ 1 ≤ i ≤ n 1 \leq i \leq n 1≤i≤n）

那么这是 m m m就需要变成 ⌈ m i ⌉ ∗ i \lceil \frac{m}{i} \rceil*i ⌈im⌉∗i。

操作次数就是 n − i + ⌈ m i ⌉ ∗ i − m n-i+\lceil \frac{m}{i} \rceil*i-m n−i+⌈im⌉∗i−m。

整理可得原式等于
( ⌈ m i ⌉ − 1 ) ∗ i + n − m (\lceil \frac{m}{i} \rceil-1)*i+n-m (⌈im⌉−1)∗i+n−m

我们把上式中的上取整转换成下取整。

( ⌊ m + i − 1 i ⌋ − 1 ) ∗ i + n − m (\lfloor \frac{m+i-1}{i} \rfloor-1)*i+n-m (⌊im+i−1⌋−1)∗i+n−m

⇉ ⌊ m − 1 i ⌋ ∗ i + n − m \rightrightarrows \lfloor \frac{m-1}{i} \rfloor*i+n-m ⇉⌊im−1⌋∗i+n−m

这时我们就可分块维护 ⌊ m − 1 i ⌋ ∗ i \lfloor \frac{m-1}{i} \rfloor*i ⌊im−1⌋∗i,记录答案最小值就行了。

代码

#include <bits/stdc++.h>
#define PI atan(1.0)*4
#define rp(i,s,t) for (register int i = (s); i <= (t); i++)
#define RP(i,t,s) for (register int i = (t); i >= (s); i--)
#define sc(x) scanf("%d",&x)
#define scl(x) scanf("%lld",&x)
#define ll long long
#define ull unsigned long long
#define mst(a,b) memset(a,b,sizeof(a))
#define lson rt<<1,l,m
#define rson rt<<1|1,m+1,r
#define pii pair<int,int>
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define m_p make_pair
#define p_b push_back
#define ins insert
#define era erase
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f3f3f3f
#define dg if(debug)
#define pY puts("YES")
#define pN puts("NO")
#define outval(a) cout << "Debuging...|" << #a << ": " << a << "\n";
#define outval2(a,b) cout << "Debuging...|" << #a << ": " << a <<"\t"<< #b << ": " << b << "\n";
#define outval3(a,b,c) cout << "Debuging...|" << #a << ": " << a <<"\t"<< #b << ": " << b <<"\t"<< #c << ": " << c << "\n";
using namespace std;
int debug = 0;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;
}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;
}
inline int read(){int s=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-') f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){s=s*10+ch-'0';ch=getchar();}return s*f;
}
const int N = 2e5+7;
int a[N];
void solve(){ll n,m;scl(n);scl(m);if(n>m) cout<<n-m<<endl;else{ll ans=inf;for(ll l=1,r;l<=n;l=r+1){r=min(n,(m-1)/((m-1)/l));ans=min(ans,(m-1)/l*l);}cout<<ans+n-m<<endl;}
}
int main(){//ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#ifdef ONLINE_JUDGE
#elsefreopen("in.txt", "r", stdin);//debug = 1;
#endif//time_t beg, end;//if(debug) beg = clock();int T=read();while(T--) solve();/*if(debug) {end = clock();printf("time:%.2fs\n", 1.0 * (end - beg) / CLOCKS_PER_SEC);}*/return 0;
}

G.Wall Game

题解

把每个坐标离散化成一个点。

对于每次占领操作，我们用并查集维护能够触碰的墙的个数。

对于一个坐标，最多能和六个相邻的坐标相连，我们枚举相邻几个个点能够触碰的墙的和 s u m sum sum以及自身产生的能够触碰的墙的个数 6 6 6。

然后用容斥的思想减去需要消去的墙的个数。

这时分为两种来考虑。

一种是自身和相邻坐标消去的墙 c n t cnt cnt，

另一种是相邻坐标中中消去的墙 m i n u s minus minus（其祖先节点不相等）。

注意需要判断相邻坐标祖先节点相同时的情况（去重），

最终该点的答案就是 s u m − 6 − c n t ∗ 2 − m i n u s ∗ 2 sum-6-cnt*2-minus*2 sum−6−cnt∗2−minus∗2。

对于每次查询操作，查询其祖先节点的答案即可。

代码

#include <bits/stdc++.h>
#define PI atan(1.0)*4
#define rp(i,s,t) for (register int i = (s); i <= (t); i++)
#define RP(i,t,s) for (register int i = (t); i >= (s); i--)
#define sc(x) scanf("%d",&x)
#define scl(x) scanf("%lld",&x)
#define ll long long
#define ull unsigned long long
#define mst(a,b) memset(a,b,sizeof(a))
#define lson rt<<1,l,m
#define rson rt<<1|1,m+1,r
#define pii pair<int,int>
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define m_p make_pair
#define p_b push_back
#define ins insert
#define era erase
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f3f3f3f
#define dg if(debug)
#define pY puts("YES")
#define pN puts("NO")
#define outval(a) cout << "Debuging...|" << #a << ": " << a << "\n";
#define outval2(a,b) cout << "Debuging...|" << #a << ": " << a <<"\t"<< #b << ": " << b << "\n";
#define outval3(a,b,c) cout << "Debuging...|" << #a << ": " << a <<"\t"<< #b << ": " << b <<"\t"<< #c << ": " << c << "\n";
using namespace std;
int debug = 0;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;
}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;
}
inline int read(){int s=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-') f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){s=s*10+ch-'0';ch=getchar();}return s*f;
}
const int N = 5e5+7;
int fa[N];
int w[N];
map<pii,int> mp;
int dir[6][2]={0,1,1,0,1,-1,0,-1,-1,0,-1,1};
int find(int x){return x==fa[x]?fa[x]:fa[x]=find(fa[x]);
}
void solve(){int n=read();int tot=0;while(n--){int op=read(),x=read(),y=read();if(op==1){mp[m_p(x,y)]=++tot;fa[tot]=tot;w[tot]=6;int sum=0;int minus=0;int cnt=0;set<int> ss;vector<pii> vv;rp(k,0,5){int nx=x+dir[k][0];int ny=y+dir[k][1];if(mp.find(m_p(nx,ny))==mp.end()) continue;int id=find(mp[m_p(nx,ny)]);vv.push_back(m_p(k,id));if(ss.find(id)!=ss.end()) cnt++;else{cnt++;sum+=w[id];ss.insert(id);}}int len=vv.size();rp(k,0,len-1){rp(j,0,len-1){if(vv[k].first==0&&vv[j].first==1&&vv[k].second!=vv[j].second) minus++;if(vv[k].first==1&&vv[j].first==2&&vv[k].second!=vv[j].second) minus++;if(vv[k].first==2&&vv[j].first==3&&vv[k].second!=vv[j].second) minus++;if(vv[k].first==3&&vv[j].first==4&&vv[k].second!=vv[j].second) minus++;if(vv[k].first==4&&vv[j].first==5&&vv[k].second!=vv[j].second) minus++;if(vv[k].first==5&&vv[j].first==0&&vv[k].second!=vv[j].second) minus++;}}for(auto val:ss){// cout<<val<<" ";fa[val]=tot;}// cout<<endl;w[tot]=sum+w[tot]-2*cnt-2*minus;// cout<<w[tot]<<endl;// outval3(sum,cnt,minus);}else{cout<<w[find(mp[m_p(x,y)])]<<endl;}}// for(auto val:mp){// pii t=val.first;// outval3(t.first,t.second,find(mp[t]));// }
}
int main(){//ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#ifdef ONLINE_JUDGE
#elsefreopen("in.txt", "r", stdin);//debug = 1;
#endif//time_t beg, end;//if(debug) beg = clock();solve();/*if(debug) {end = clock();printf("time:%.2fs\n", 1.0 * (end - beg) / CLOCKS_PER_SEC);}*/return 0;
}

J.Grammy and Jewelry

题解

算是比较套路的题。

根据贪心的原则，我们每次取一个宝珠后应该立马返回到原点。

而我们到这个宝珠的位置走的路和返回时走的路应该是最短路。

转换一下就是我们要取一个价值为 a [ i ] a[i] a[i]的宝珠，需要消耗容量为 2 ∗ d i s [ i ] 2*dis[i] 2∗dis[i]的时间（ d i s [ i ] dis[i] dis[i]表示从 1 1 1到 i i i的最短路）。

这样就转换成了经典的完全背包问题。

关于背包问题不懂的可以看经典的背包九讲。

代码

#include <bits/stdc++.h>
#define _for(i, a) for(int i = 0, le = (a); i < le; ++i)
#define _rep(i, a, b) for(int i = (a), le = (b); i <= le; ++i)
typedef long long LL;
#define INF 0x3f3f3f3f
const int N = 3007;
using namespace std;LL read() {LL x(0), f(1); char c = getchar();while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}while(c >= '0' && c <= '9') {x = x * 10 + c - '0';c = getchar();}return x * f;
}
vector<int> G[N];
int a[N];
int dis[N];
int vis[N];
int dp[N];
struct node{int u,w;friend bool operator <(node a,node b){return a.w>b.w;}
};
int n,m,T;
void Dijkstra(){priority_queue<node> q;_rep(i,1,n) dis[i]=0x3f3f3f3f;q.push(node{1,0});dis[1]=0;memset(vis,0,sizeof vis);while(!q.empty()){node t=q.top();q.pop();vis[t.u]=1;int u=t.u;int w=t.w;// cout<<u<<" "<<w<<endl;for(auto v:G[u]){// cout<<v<<" "<<dis[v]<<endl;if(!vis[v]&&dis[v]>w+1){dis[v]=w+1;q.push(node{v,dis[v]});}}}
}
int main() {// freopen("in.txt", "r", stdin);n=read(),m=read(),T=read();_rep(i,2,n) a[i]=read();_rep(i,1,m){int u=read(),v=read();if(u==v) continue;G[u].push_back(v);G[v].push_back(u);}// for(int u=1;u<=n;u++){//     cout<<u<<endl;//     for(auto v:G[u]) cout<<v<<" ";//     cout<<endl;// }Dijkstra();// _rep(i, 1, n) printf("dis[%d]:%d\n", i, dis[i]);_rep(i, 1, n) {for(int j = dis[i] * 2; j <=T ; ++j) {dp[j] = max(dp[j], dp[j - dis[i] * 2] + a[i]);}}_rep(i, 1, T) printf("%d%s", dp[i], (i == T ? "\n":" "));return 0;
}

L.String Freshman

题解

c f cf cf上的题意有点错误，给的字符串应该是匹配串。

因此我们判断一下是否存在前缀子串的最长公共前后缀长度不为 0 0 0（即next数组不为 0 0 0）。

更简单的做法就是判断是否有字符和第一个字符相等即可。

代码

#include <bits/stdc++.h>
#define PI atan(1.0)*4
#define rp(i,s,t) for (register int i = (s); i <= (t); i++)
#define RP(i,t,s) for (register int i = (t); i >= (s); i--)
#define sc(x) scanf("%d",&x)
#define scl(x) scanf("%lld",&x)
#define ll long long
#define ull unsigned long long
#define mst(a,b) memset(a,b,sizeof(a))
#define lson rt<<1,l,m
#define rson rt<<1|1,m+1,r
#define pii pair<int,int>
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define m_p make_pair
#define p_b push_back
#define ins insert
#define era erase
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f3f3f3f
#define dg if(debug)
#define pY puts("YES")
#define pN puts("NO")
#define outval(a) cout << "Debuging...|" << #a << ": " << a << "\n";
#define outval2(a,b) cout << "Debuging...|" << #a << ": " << a <<"\t"<< #b << ": " << b << "\n";
#define outval3(a,b,c) cout << "Debuging...|" << #a << ": " << a <<"\t"<< #b << ": " << b <<"\t"<< #c << ": " << c << "\n";
using namespace std;
int debug = 0;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;
}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;
}
inline int read(){int s=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-') f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){s=s*10+ch-'0';ch=getchar();}return s*f;
}
const int N = 2e5+7;
int a[N];
void solve(){int n=read();string s;cin>>s;rp(i,1,n-1){if(s[i]==s[0]){printf("Wrong Answer\n");return ;}}printf("Correct\n");
}
int main(){//ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#ifdef ONLINE_JUDGE
#elsefreopen("in.txt", "r", stdin);//debug = 1;
#endif//time_t beg, end;//if(debug) beg = clock();solve();/*if(debug) {end = clock();printf("time:%.2fs\n", 1.0 * (end - beg) / CLOCKS_PER_SEC);}*/return 0;
}

M.Game Theory

题解

假结论题。

先说结论：输出 0.0000 0.0000 0.0000即可。

当时感觉题意有点模糊，后面读懂题意后验证了一下就交了。

首先我们知道每个学生是不知道 G r a m m y Grammy Grammy的选择，而又因为学生每次都是最优的选择，因此我们可以先算出一个学生在 [ 1 , 20 ] [1,20] [1,20]选择一个数后赢的分数的期望，每次都选择这 20 20 20个数中期望最大的即可。

我们写出计算期望的代码可以发现，每个数赢的分数的期望都是 0 0 0。

而学生和 G r a m m y Grammy Grammy的分数又是互补的，只有一个输了，另一个才能赢。

因此 G r a m m y Grammy Grammy赢的分数的期望也是 0 0 0。

代码

#include <bits/stdc++.h>
#define _for(i, a) for(int i = 0, le = (a); i < le; ++i)
#define _rep(i, a, b) for(int i = (a), le = (b); i <= le; ++i)
typedef long long LL;
using namespace std;
LL read() {LL x(0), f(1); char c = getchar();while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}while(c >= '0' && c <= '9') {x = x * 10 + c - '0';c = getchar();}return x * f;
}
const int maxn = 100005;int getSco(int G, int S) {if(G > S) return G - S - 10;if(S > G) return G - S + 10;return G - S;
}double sco(int x) {double ans = 0;_rep(i, 1, 20) {ans += getSco(i, x);}return ans / 20.0;
}int main() {// _rep(i, 1, 20) {//     printf("i:%d\tans:%.10f\n", i, sco(i));// }int x = read();printf("0.0000\n");return 0;
}