C++奥赛一本通刷题记录(递推)

2017.11.8 By gwj1139177410


#include<iostream>

using namespace std;
const int maxn=1000010, mod=1000;
int f[maxn];
int main(){
   f[1] = f[2] = 1;
   for(int i = 3; i <= maxn; i++)
       f[i] = (f[i-1]+f[i-2])%mod;//bugs
   int t;  cin>>t;
   while(t--){
       int n;  cin>>n;
       cout<<f[n]<<"
";
   }
   return 0;
}

pell数列 noioj1071&openjudge1788


#include<iostream>

using namespace std;
const int maxn = 1000000+10,mod=32767;
int f[maxn];
int main(){
   f[1]=1; f[2]=2;
   int k = maxn;
   for(int i = 3; i <= k; i++)f[i]=(2*f[i-1]+f[i-2])%mod;//bugs
   int n;  cin>>n;
   while(n--){
       cin>>k;
       cout<<f[k]<<"
";
   }
   return 0;
}

上台阶 openjudge3525

//f[i]表示走i阶台阶的走法数目
//因为每次只能走一阶或者两阶,所以由f[i-1]和f[i-2]相加转移而来

#include<iostream>

using namespace std;
const int maxn = 110;
int f[maxn], n;
int main(){
   f[1] = 1; f[2] = 2; f[3] = 4;
   for(int i = 4; i < maxn; i++)f[i] = f[i-1]+f[i-2]+f[i-3];
   while(cin>>n && n)cout<<f[n]<<"
";
   return 0;
}

流感传染 openjudge6262

//BFS模板

#include<iostream>


#include<queue>

using namespace std;
const int maxn = 110;

char a[maxn][maxn];
int n, m, cnt;
int vis[maxn][maxn];

struct node{
   int x, y, step;
node(int x,int y, int step){
       this->x = x;
       this->y = y;
       this->step = step;
   };
};
const int dx[] = {0,-1,0,1};
const int dy[] = {1,0,-1,0};
queue<node>q;
void bfs(){
   while(q.size()) {
       node t = q.front(); q.pop();
       for(int i = 0; i < 4; i++){
           int nx = t.x+dx[i], ny = t.y+dy[i], st = t.step+1;
           if(st == m){
               cout<<cnt<<"
";
               return ;
           }
           if(nx>=1&&nx<=n&&ny>=1&&ny<=n&&a[nx][ny]!='#'&&!vis[nx][ny]){
               q.push(node(nx,ny,st));
               vis[nx][ny] = 1;
               cnt++;
           }
       }
   }
   cout<<cnt<<"
";//bugs
   return ;
}

int main(){
   cin>>n;  cin.get(); //datain bugs
   for(int i = 1; i <= n; i++){
       for(int j = 1; j <= n; j++){
           a[i][j]=getchar();
           if(a[i][j]=='@'){
               q.push(node(i,j,0));
               vis[i][j] = 1;
               cnt++;
           }
       }
       getchar();
   }
   cin>>m; cin.get();
   bfs();
   return 0;
}

放苹果 openjudge666&POJ1664&luogu2386

//f[i][j]表示i个苹果j个盘子的放法数目
//j>i时,去掉空盘不影响结果; j<=i时,对盘子是否空着分类讨论;*

#include<iostream>

using namespace std;
const int maxn = 11;
int f[maxn][maxn];
int main(){
   for(int i = 0; i < maxn; i++)f[0][i]=f[i][1]=1;
   for(int i = 1; i < maxn; i++)//pojbugs
       for(int j = 2; j < maxn; j++)
           f[i][j] = j>i?f[i][i]:f[i][j-1]+f[i-j][j];//所有盘子又有苹果时每个盘子都去掉一个苹果不影响结果
   int t; cin>>t;
   while(t--){
       int n, m;
       cin>>n>>m;
       cout<<f[n][m]<<"
";
   }
   return 0;
}

吃糖果 openjudge1944

//f[i]表示名名吃i块巧克力的方案数, f[0]=f[1]=1;

#include<iostream>

using namespace std;
int f[30];
int main(){
   int n;  cin>>n;
   f[0] = f[1] = 1;
   for(int i = 2; i <= n; i++)
       f[i%2]=f[(i-1)%2]+f[(i-2)%2];
   cout<<f[n%2];
   return 0;
}

移动路线 openjudge2781

//f[i][j]表示从(m,1)到(i,j)的不同路线数目

#include<iostream>

using namespace std;
const int maxn = 30;
int f[maxn][maxn];
int main(){
   int m, n;  cin>>m>>n;
   f[m][0] = 1;
   for(int i = m; i >= 1; i--)
       for(int j = 1; j <= n; j++)
           f[i][j] = f[i+1][j]+f[i][j-1];
   cout<<f[1][n];
   return 0;
}

判断整除 openjudge3531

//f[i][j]表示用前i个数计算能得到余数j*

#include<iostream>

using namespace std;
const int maxn = 10010, maxk = 110;
int a[maxn], f[maxn][maxk];
int main(){
   int n, k;  cin>>n>>k;
   for(int i = 1; i <= n; i++)cin>>a[i],a[i]%=k;
   f[1][a[1]] = 1;
   for(int i = 2; i <= n; i++)
       for(int j = 0; j < k; j++)
           if(f[i-1][j])f[i][(j+a[i])%k]=f[i][(j-a[i]+k)%k]=1;
   if(f[n][0])cout<<"YES
";else cout<<"NO
";
   return 0;
}

踩方格 openjudge4982

//l[i],r[i],u[i]分别表示最后一步向左向右向上走到第i格*

#include<iostream>

using namespace std;
const int maxn = 30;
int l[maxn], r[maxn], u[maxn];
int main(){
   int n;  cin>>n;
   l[1] = r[1] = u[1] = 1;
   for(int i = 2; i <= n; i++){
       l[i] = l[i-1]+u[i-1];
       r[i] = r[i-1]+u[i-1];
       u[i] = l[i-1]+r[i-1]+u[i-1];
   }
   cout<<l[n]+r[n]+u[n]<<"
";
   return 0;
}

山区建小学 openjudge7624

//f[i][j]表示1..i中建j个小学的最小距离和.(这里的j可以看成是最后一所学校管辖区的终点)
//f[i][j]=min(f[i][j],f[k][j-1]+s[k+1][i]),j-1<=k<i;

#include<iostream>


#include<algorithm>


#include<cmath>

using namespace std;
const int maxn = 510;
int a[maxn],dis[maxn][maxn],s[maxn][maxn],f[maxn][maxn];
//s[管辖区起点][管辖区终点]=这片辖区内建一个学校，区内村庄到学校的最小距离和
//一个结论:因为i..j中选一个点使所有点到这个点的总距离最小，这个点一定在中点位置（反证法，左移右移时）
int dist(int l, int r){
   int m = (l+r)/2, sum = 0;
   for(int i = l; i <= r; i++)sum += dis[i][m];
   return sum;
}
int main(){
   int m, n;
   cin>>m>>n;
   for(int i = 2; i <= m; i++)cin>>a[i],a[i]+=a[i-1];
   //初始化两两距离
   for(int i = 1; i <= m; i++)
       for(int j = 1; j <= m; j++)
           dis[i][j] = i==j?0:abs(a[j]-a[i]);
   //计算一个管辖从i到j村庄的学校到这些村庄的距离和
   for(int i = 1; i <= m; i++)
       for(int j = 1; j <= m; j++)
           s[i][j] = dist(i,j);
   //初始化
   for(int i = 1; i <= m; i++)
       for(int j = 1; j <= m; j++)
           f[i][j] = i==j?0:0xfffff;
   for(int i = 1; i <= m; i++)f[i][1] = s[1][i];//只建一所学校
   //DP
   for(int i = 2; i <= m; i++)//村庄
       for(int j = 2; j <= min(i,n); j++)//学校
           for(int k = j-1; k < i; k++)//枚举已有的(最后一所)学校管辖的范围(终点)
               if(i!=j)f[i][j] = min(f[i][j],f[k][j-1]+s[k+1][i]);
   cout<<f[m][n];
   return 0;
}