Summer training #8
Posted Aragaki
tags:
篇首语：本文由小常识网(cha138.com)小编为大家整理，主要介绍了Summer training #8相关的知识，希望对你有一定的参考价值。
A：
B：按题意直接暴力找符合题意的数的个数
#include <bits/stdc++.h> #include <cstring> #include <iostream> #include <algorithm> #define foror(i,a,b) for(i=a;i<b;i++) #define foror2(i,a,b) for(i=a;i>b;i--) #define EPS 1.0e-6 #define PI acos(-1.0) #define INF 3000000000 #define MOD 1000000009 #define mem(a,b) memset((a),b,sizeof(a)) #define TS printf("!!!\n") #define lson o<<1, l, m #define rson o<<1|1, m+1, r //using ll = long long; //using ull= unsigned long long; //std::ios::sync_with_stdio(false); using namespace std; //priority_queue<int,vector<int>,greater<int>> que; typedef long long ll; int num[15][2];//+ 1 - 2 * 3 / 4 int main() { //freopen("in.txt", "r", stdin); //freopen("out.txt", "w", stdout); int n; string a; int ans=0; cin >> n; int now; for(int i=1;i<=n;i++) { cin >> a >> now; num[i][1]=now; if(a[0]==‘A‘) num[i][0]=1; if(a[0]==‘S‘) num[i][0]=2; if(a[0]==‘M‘) num[i][0]=3; if(a[0]==‘D‘) num[i][0]=4; } for(int i=1;i<=100;i++) { int cur=i; for(int j=1;j<=n;j++) { if(num[j][0]==1) { cur+=num[j][1]; } if(num[j][0]==2) { cur-=num[j][1]; if(cur<0) { ans++; break; } } if(num[j][0]==3) { cur*=num[j][1]; } if(num[j][0]==4) { if(cur%num[j][1]!=0) { ans++; break; } else cur/=num[j][1]; } } } cout<<ans<<endl; return 0; }
C：水
D：s[i][j]可以向四个方向跳s[i][j]个格子 问能不能从左上角跳到右下角 BFS
#include <bits/stdc++.h> #include <cstring> #include <iostream> #include <algorithm> #define foror(i,a,b) for(i=a;i<b;i++) #define foror2(i,a,b) for(i=a;i>b;i--) #define EPS 1.0e-6 #define PI acos(-1.0) #define INF 3000000000 #define MOD 1000000009 #define mem(a,b) memset((a),b,sizeof(a)) #define TS printf("!!!\n") #define lson o<<1, l, m #define rson o<<1|1, m+1, r //using ll = long long; //using ull= unsigned long long; //std::ios::sync_with_stdio(false); using namespace std; //priority_queue<int,vector<int>,greater<int>> que; typedef long long ll; int dir[5][2]; int m,n; char s1[505][505]; int a[505][505]; int ans[505][505]; int flag=0; pair<int,int> now; queue<pair<int,int> > que; void bfs() { for(int i=0;i<=500;i++) for(int j=0;j<=500;j++) ans[i][j]=2e9; ans[1][1]=0; int dx,dy; while(!que.empty()) { now=que.front(); que.pop(); int f=now.first; int s=now.second; int j=a[f][s]; for(int i=0;i<4;i++) { dx=f+j*dir[i][0]; dy=s+dir[i][1]*j; if(dx>=1&&dx<=m&&dy>=1&&dy<=n&&ans[dx][dy]>ans[f][s]+1) { ans[dx][dy]=ans[f][s]+1; now.first=dx,now.second=dy; que.push(now); } } } } int main() { //freopen("in.txt", "r", stdin); //freopen("out.txt", "w", stdout); cin >> m >> n; dir[0][0]=-1,dir[0][1]=0; dir[1][0]=1,dir[1][1]=0; dir[2][0]=0,dir[2][1]=1; dir[3][0]=0,dir[3][1]=-1; for(int i=1;i<=m;i++) { scanf("%s",s1[i]+1); for(int j=1;j<=n;j++) a[i][j]=s1[i][j]-‘0‘; } /*for(int i=0;i<m;i++) { for(int j=0;j<n;j++) printf("%d",a[i][j]); cout<<endl; }*/ //TS; now.first=now.second=1; que.push(now); bfs(); //TS; if(ans[m][n]==2e9) cout<<"IMPOSSIBLE"<<endl; else cout<<ans[m][n]<<endl; return 0; }
E：水
F：水
G：水
H：加权区间调度问题 瞎DP
#include <bits/stdc++.h> #include <cstring> #include <iostream> #include <algorithm> #define foror(i,a,b) for(i=a;i<b;i++) #define foror2(i,a,b) for(i=a;i>b;i--) #define EPS 1.0e-6 #define PI acos(-1.0) #define INF 3000000000 #define MOD 1000000009 #define mem(a,b) memset((a),b,sizeof(a)) #define TS printf("!!!\n") #define lson o<<1, l, m #define rson o<<1|1, m+1, r //using ll = long long; //using ull= unsigned long long; //std::ios::sync_with_stdio(false); using namespace std; //priority_queue<int,vector<int>,greater<int>> que; typedef long long ll; struct wave { ll start; ll w; ll t; ll end; }now[300005]; bool cmp(wave a,wave b) { return a.start>b.start; } ll dp[1000010]; int main() { //freopen("in.txt", "r", stdin); //freopen("out.txt", "w", stdout); int n; ll maxn=0; cin >> n; for(int i=1;i<=n;i++) { int t; scanf("%d %d %d",&now[i].start,&now[i].w,&now[i].t); now[i].end=now[i].t+now[i].start; maxn=max(maxn,now[i].end); } sort(now+1,now+1+n,cmp); int pop=1; for(int i=1000000;i>=1;i--) { if(i==now[pop].start) { if(i+now[pop].t>1000000) dp[i]=max(dp[i+1],now[pop].w); else dp[i]=max(dp[i+1],dp[i+now[pop].t]+now[pop].w); pop++; } else dp[i]=dp[i+1]; } cout<<dp[1]<<endl; return 0; }
J：每个格子的值为其为中心九宫格之和 问最终的值 其实一起都要除九的话就不用除了
#include <bits/stdc++.h> #include <cstring> #include <iostream> #include <algorithm> #define foror(i,a,b) for(i=a;i<b;i++) #define foror2(i,a,b) for(i=a;i>b;i--) #define EPS 1.0e-6 #define PI acos(-1.0) #define INF 3000000000 #define MOD 1000000009 #define mem(a,b) memset((a),b,sizeof(a)) #define TS printf("!!!\n") #define lson o<<1, l, m #define rson o<<1|1, m+1, r //using ll = long long; //using ull= unsigned long long; //std::ios::sync_with_stdio(false); using namespace std; //priority_queue<int,vector<int>,greater<int>> que; typedef long long ll; int now[110][110]; int doit[110][110]; set<int> ans; int main() { //freopen("in.txt", "r", stdin); //freopen("out.txt", "w", stdout); int m,n,time; cin >> m >> n >> time; for(int i=0;i<n;i++) for(int j=0;j<m;j++) scanf("%d",&now[i][j]); while(time--) { mem(doit,0); for(int i=0;i<n;i++) { for(int j=0;j<m;j++) { for(int x=-1;x<=1;x++) { for(int y=-1;y<=1;y++) { int dx=(n+i+x)%n; int dy=(m+j+y)%m; doit[i][j]+=now[dx][dy]; } } } } for(int i=0;i<n;i++) for(int j=0;j<m;j++) now[i][j]=doit[i][j]; } for(int i=0;i<n;i++) for(int j=0;j<m;j++) ans.insert(now[i][j]); cout<<ans.size()<<endl; return 0; }
K：DFS+回溯 注意！！！每个轮子只能有一个旋转状态 如果发生矛盾了就转不动了！
#include <bits/stdc++.h> #include <cstring> #include <iostream> #include <algorithm> #define foror(i,a,b) for(i=a;i<b;i++) #define foror2(i,a,b) for(i=a;i>b;i--) #define EPS 1.0e-6 #define PI acos(-1.0) #define INF 3000000000 #define MOD 1000000009 #define mem(a,b) memset((a),b,sizeof(a)) #define TS printf("!!!\n") #define lson o<<1, l, m #define rson o<<1|1, m+1, r //using ll = long long; //using ull= unsigned long long; //std::ios::sync_with_stdio(false); using namespace std; //priority_queue<int,vector<int>,greater<int>> que; typedef long long ll; int n; int flag=-1; struct point { int x,y,r; }a[1005]; vector<int> v[1005]; int visit[1005]; int ans=0; int anser; bool check(point a,point b) { int dx=abs(a.x-b.x); int dy=abs(a.y-b.y); int dr=a.r+b.r; if(dx*dx+dy*dy==dr*dr) return true; return false; } void dfs(int num,int now) { if(flag==1) return ; for(int i=0;i<v[num].size();i++) { int next=v[num][i]; if(flag==1) return ; if(visit[next]==-1) { visit[next]=now; dfs(next,now^1); } if(visit[next]!=now) { flag=1; return ; } } } int gcd(int x,int y){return y==0?x:gcd(y,x%y);} int main() { //freopen("in.txt", "r", stdin); //freopen("out.txt", "w", stdout); cin >> n; mem(visit,-1); visit[1]=0; for(int i=1;i<=n;i++) { cin >> a[i].x >> a[i].y >> a[i].r; } for(int i=1;i<n;i++) for(int j=i+1;j<=n;j++) { if(check(a[i],a[j])) { v[i].push_back(j); v[j].push_back(i); } } dfs(1,1); if(flag==1) { cout<<"The input gear cannot move."<<endl; return 0; } if(visit[n]==-1) { cout<<"The input gear is not connected to the output gear."<<endl; return 0; } int chu=gcd(a[1].r,a[n].r); if(visit[1]!=visit[n]) cout<<"-"; printf("%d:%d\n",a[1].r/chu,a[n].r/chu); return 0; }
以上是关于Summer training #8的主要内容，如果未能解决你的问题，请参考以下文章
2017ecjtu-summer training #5 UVA10382