Summer training #8

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;
}

View Code

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;
}

View Code

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;
}

View Code

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;
}

View Code

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;
}

View Code

posted @ 2017-07-17 22:20 Aragaki 阅读(168) 评论(0) 编辑收藏举报

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