title: C++奥赛一本通刷题记录(递推)
date: 2017-11-08
tags:
- 一本通
- openjudege
categories: OI
C++奥赛一本通刷题记录(递推)
2017.11.8 By gwj1139177410
- 斐波那契数列 openjudge1760
#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]<<"\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]<<"\n";
}
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]<<"\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<<"\n";
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<<"\n";//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]<<"\n";
}
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\n";else cout<<"NO\n";
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]<<"\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;
}