题解

294 条题解

  • 0
    @ 2016-07-25 18:21:21
    #include <stdio.h>
    #include <stdlib.h>
    int cx,cy,m,n;
    int map[16][16];
    int f(int a,int b)
    {
        if(map[a][b]==1)
            return 0;
        if(a==0&&b==0)
            return 1;
        if(a==0)
            return f(a,b-1);
        if(b==0)
            return f(a-1,b);
        return f(a-1,b)+f(a,b-1);
    }
    int main()
    {
        int paths;
        const int x[10]={ 2, 1,-2,-1, 2, 1,-2,-1, 2,-2};
        const int y[10]={ 1, 2,-1,-2,-1,-2, 1, 2, 2,-2};
        scanf("%d %d %d %d",&n,&m,&cx,&cy);
        {
            int i;
            map[cx][cy]=1;
            for(i=0;i<10;i++)
                if(cx+x[i]>=0&&cx+x[i]<=n&&cy+y[i]>=0&&cy+y[i]<=m)
                    map[cx+x[i]][cy+y[i]]=1;
        }
        paths=f(n,m);
        printf("%d",paths);
        return 0;
    }
    
  • 0
    @ 2016-07-13 22:13:50
    记录信息
    评测状态    Accepted
    题目  P1121 马拦过河卒
    递交时间    2016-05-17 21:50:30
    代码语言    C++
    评测机 ShadowShore
    消耗时间    0 ms
    消耗内存    656 KiB
    评测时间    2016-05-17 21:50:32
    评测结果
    编译成功
    
    测试数据 #0: Accepted, time = 0 ms, mem = 652 KiB, score = 10
    测试数据 #1: Accepted, time = 0 ms, mem = 656 KiB, score = 10
    测试数据 #2: Accepted, time = 0 ms, mem = 652 KiB, score = 10
    测试数据 #3: Accepted, time = 0 ms, mem = 652 KiB, score = 10
    测试数据 #4: Accepted, time = 0 ms, mem = 656 KiB, score = 10
    测试数据 #5: Accepted, time = 0 ms, mem = 656 KiB, score = 10
    测试数据 #6: Accepted, time = 0 ms, mem = 656 KiB, score = 10
    测试数据 #7: Accepted, time = 0 ms, mem = 652 KiB, score = 10
    测试数据 #8: Accepted, time = 0 ms, mem = 656 KiB, score = 10
    测试数据 #9: Accepted, time = 0 ms, mem = 656 KiB, score = 10
    Accepted, time = 0 ms, mem = 656 KiB, score = 100
    代码
    #include<cstdio>
    #include<cstring>
    using namespace std;
    int main()
    {
      long n,m,x,y;
      int l1[10]={0,0,1,-1,1,-1,-2,-2,2,2};
      int l2[10]={0,0,-2,-2,2,2,-1,1,1,-1};
      long long a[101][101],b[101][101];
      memset(a,0,sizeof(a));
      memset(b,0,sizeof(b));
      scanf("%ld %ld %ld %ld",&n,&m,&x,&y);
      for(int i=1;i<=9;i++)
        if(y+l2[i]>=0 && x+l1[i]>=0)
          b[x+l1[i]][y+l2[i]]=1;
      a[0][0]=1;
      for(int i=1;i<=m;i++)
        if(b[0][i]==0)
          a[0][i]=a[0][i-1];
      for(int i=1;i<=n;i++)
        if(b[i][0]==0)
          a[i][0]=a[i-1][0];
      for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
          if(b[i][j]==0)
            a[i][j]=a[i-1][j]+a[i][j-1];
      printf("%I64d",a[n][m]);
      return 0;
    }
    
  • 0
    @ 2016-07-12 10:43:56

    #include<iostream>
    using namespace std;
    int main()
    {
    int x,y,s,t;
    cin>>x>>y>>s>>t;
    long long a[x+1][y+1],b[x+1][y+1];
    for(int i=0;i<=x;i++)
    for(int j=0;j<=y;j++)
    {
    a[i][j]=0;
    b[i][j]=0;
    }
    a[0][0]=1;
    b[s+1][t+2]=1;
    b[s+1][t-2]=1;
    b[s+2][t+1]=1;
    b[s+2][t-1]=1;
    b[s-1][t-2]=1;
    b[s-1][t+2]=1;
    b[s-2][t+1]=1;
    b[s-2][t-1]=1;
    b[s][t]=1;
    for(int i=0;i<=x;i++)
    for(int j=0;j<=y;j++)
    {
    if(i>=1&&b[i][j]==0)
    a[i][j]+=a[i-1][j];
    if(j>=1&&b[i][j]==0)
    a[i][j]+=a[i][j-1];
    }
    cout<<a[x][y];
    }

  • 0
    @ 2016-04-03 11:07:14

    #include<iostream>
    #include<cstring>
    using namespace std;
    void chu();
    void chick(int n,int m);
    void machick(int n, int m);
    int getabc();
    struct Nzb
    {
    int x;
    int y;
    };
    Nzb ma,zhong;
    int a[20][20];
    bool malan[20][20];
    int main()
    {
    cin >> zhong.x >> zhong.y >> ma.x >> ma.y;
    memset(malan, true, sizeof(malan));
    machick(ma.x, ma.y);
    chu();
    cout << getabc()<<endl;
    return 0;
    }
    void chu()
    {
    int Temp1,Temp2;
    memset(a, 0, sizeof(a));
    for (Temp1 = 0; Temp1 <= zhong.x; Temp1++)
    {
    if (!malan[Temp1][0])
    {
    a[Temp1][0] = 0;
    break;
    }
    else
    a[Temp1][0] = 1;
    }
    for (Temp2 = 0; Temp2 <= zhong.y; Temp2++)
    {
    if (!malan[0][Temp2])
    {
    a[0][Temp2] = 0;
    break;
    }
    else
    a[0][Temp2] = 1;
    }
    }
    void machick(int n, int m)
    {
    if ((n + 1 >= 0) && (n + 1 <= 15) && (m + 2 >= 0)&&(m + 2 <= 15))
    malan[n + 1][m + 2] = false;
    if ((n - 1 >= 0) && (n - 1 <= 15) && (m + 2 >= 0) && (m + 2 <= 15))
    malan[n - 1][m + 2] = false;
    if ((n + 1 >= 0) && (n + 1 <= 15) && (m - 2 >= 0) && (m - 2 <= 15))
    malan[n + 1][m - 2] = false;
    if ((n - 1 >= 0) && (n - 1 <= 15) && (m - 2 >= 0) && (m - 2 <= 15))
    malan[n - 1][m - 2] = false;
    if ((n + 2 >= 0) && (n + 2 <= 15) && (m + 1 >= 0) && (m + 1 <= 15))
    malan[n + 2][m + 1] = false;
    if ((n - 2 >= 0) && (n - 2 <= 15) && (m + 1 >= 0) && (m + 1 <= 15))
    malan[n - 2][m + 1] = false;
    if ((n + 2 >= 0) && (n + 2 <= 15) && (m - 1 >= 0) && (m - 1 <= 15))
    malan[n + 2][m - 1] = false;
    if ((n - 2 >= 0) && (n - 2 <= 15) && (m - 1 >= 0) && (m -1 <= 15))
    malan[n - 2][m - 1] = false;
    malan[n][m] = false;
    }
    int getabc()
    {
    int Temp1, Temp2;
    for (Temp1 = 1; Temp1 <= zhong.x; Temp1++)
    {
    for (Temp2 = 1; Temp2 <= zhong.y; Temp2++)
    {
    if (malan[Temp1][Temp2])
    {
    if (malan[Temp1 - 1][Temp2])
    a[Temp1][Temp2] += a[Temp1 - 1][Temp2];
    if (malan[Temp1][Temp2 - 1])
    a[Temp1][Temp2] += a[Temp1][Temp2 - 1];
    }
    }
    }
    return a[zhong.x][zhong.y];
    }

  • 0
    @ 2015-11-05 22:24:04

    #include<iostream>
    #include<cmath>
    #include<cstring>
    using namespace std;

    int martix[10000][10000];

    int main(){
    int a,b,c,d;
    cin>>a>>b>>c>>d;
    memset(martix,0,sizeof(martix));
    martix[c][d]=-1;
    for(int i=0;i<=a;i++)
    martix[i][0]=1;
    for(int i=0;i<=b;i++)
    martix[0][i]=1;
    if(c+2<=10000){
    if(d+1>=0 && d+1<10000) martix[c+2][d+1]=-1;
    if(d-1>=0 && d-1<10000) martix[c+2][d-1]=-1;
    }
    if(c-2<=10000 && c>=2){
    if(d+1>=0 && d+1<10000) martix[c-2][d+1]=-1;
    if(d-1>=0 && d-1<10000) martix[c-2][d-1]=-1;
    }
    if(d+2<=10000){
    if(c+1>=0 && c+1<10000) martix[c+1][d+2]=-1;
    if(c-1>=0 && c-1<10000) martix[c-1][d+2]=-1;
    }
    if(d-2<=10000 && d>=2){
    if(c+1>=0 && c+1<10000) martix[c+1][d-2]=-1;
    if(c-1>=0 && c-1<10000) martix[c-1][d-2]=-1;
    }
    if(c==0 || d==0){ martix[c][d]=-1; }
    for(int i=0;i<=a;i++){
    if(martix[i][0]==-1){
    for(int j=i+1;j<=a;j++){
    martix[j][0]=0;
    }
    break;
    }
    }
    for(int i=0;i<=b;i++){
    if(martix[0][i]==-1){
    for(int j=i+1;j<=b;j++){
    martix[0][j]=0;
    }
    break;
    }
    }
    for(int i=1;i<=a;i++){
    for(int j=1;j<=b;j++){
    if(martix[i][j]==-1) continue;
    else if(martix[i-1][j]!=-1 && martix[i][j-1]!=-1) martix[i][j]=martix[i-1][j]+martix[i][j-1];
    else if(martix[i-1][j]==-1 && martix[i][j-1]!=-1) martix[i][j]=martix[i][j-1];
    else if(martix[i-1][j]!=-1 && martix[i][j-1]==-1) martix[i][j]=martix[i-1][j];
    else if(martix[i-1][j]==-1 && martix[i][j-1]==-1) continue;
    }
    }
    cout<<(martix[a][b]==-1?0:martix[a][b]);
    return 0;
    }

  • 0
    @ 2015-10-31 15:46:57

    var a,f:array[-10..100,-10..100] of longint;
    i,j,k,l,n,m,x,y:longint;
    begin
    read(n,m,x,y);
    a[x,y]:=1;
    a[x-1,y-2]:=1;
    a[x-1,y+2]:=1;
    a[x-2,y-1]:=1;
    a[x-2,y+1]:=1;
    a[x+1,y+2]:=1;
    a[x+1,y-2]:=1;
    a[x+2,y-1]:=1;
    a[x+2,y+1]:=1;
    f[0,-1]:=1;
    for i:=0 to n do
    for j:=0 to m do
    if a[i,j]<>1 then
    f[i,j]:=f[i-1,j]+f[i,j-1];
    writeln(f[n,m]);
    end.
    f[i,j]表示到(i,j)的方案数,然后知道只能由(i-1,j)和(i,j-1)到达,所以方案数也是由这两个点得到。然后。。没有然后了。

  • 0
    @ 2015-10-05 09:29:02

    此题水炸天【暴力搜索本来只想骗分也能过】
    var a:array[0..20,0..20]of boolean;i,j,x,y,m,n,ans:longint;

    procedure zhunbei;
    begin
    fillchar(a,sizeof(a),true);
    a[x,y]:=false;
    if x-2>=0 then begin {马的上方}
    if y-1>=0 then a[x-2,y-1]:=false;
    if y+1<=m then a[x-2,y+1]:=false;
    end;
    if x+2<=n then begin {马的下方}
    if y-1>=0 then a[x+2,y-1]:=false;
    if y+1<=m then a[x+2,y+1]:=false;
    end;
    if y-2>=0 then begin {马的左方}
    if x-1>=0 then a[x-1,y-2]:=false;
    if x+1<=n then a[x+1,y-2]:=false;
    end;
    if y+2<=m then begin {马的右方}
    if x-1>=0 then a[x-1,y+2]:=false;
    if x+1<=n then a[x+1,y+2]:=false;
    end;
    end;

    procedure dfs(i,j:longint);
    begin
    if not(a[i,j]) then exit;
    if (i=n) and (j=m) then begin
    inc(ans); exit; end;
    if i<n then dfs(i+1,j);
    if j<m then dfs(i,j+1);
    end;

    begin
    read(n,m,x,y);
    zhunbei;
    ans:=0;
    dfs(0,0);
    writeln(ans);
    end.

  • 0
    @ 2015-09-20 16:45:22

    题解:http://www.cnblogs.com/xtx1999/p/4823720.html
    有疑问请留言 共同进步

  • 0
    @ 2015-09-20 11:33:31

    编译成功

    测试数据 #0: Accepted, time = 0 ms, mem = 272 KiB, score = 10
    测试数据 #1: Accepted, time = 0 ms, mem = 272 KiB, score = 10
    测试数据 #2: Accepted, time = 0 ms, mem = 276 KiB, score = 10
    测试数据 #3: Accepted, time = 0 ms, mem = 276 KiB, score = 10
    测试数据 #4: Accepted, time = 0 ms, mem = 272 KiB, score = 10
    测试数据 #5: Accepted, time = 0 ms, mem = 272 KiB, score = 10
    测试数据 #6: Accepted, time = 0 ms, mem = 276 KiB, score = 10
    测试数据 #7: Accepted, time = 0 ms, mem = 276 KiB, score = 10
    测试数据 #8: Accepted, time = 0 ms, mem = 272 KiB, score = 10
    测试数据 #9: Accepted, time = 0 ms, mem = 276 KiB, score = 10
    Accepted, time = 0 ms, mem = 276 KiB, score = 100
    #include<iostream>
    #include<cstdio>
    using namespace std;
    int x1,y1,x2,y2;
    int f[16][16],a[16][16],ans;
    int mx[8]={1,2,2,1,-1,-2,-2,-1};
    int my[8]={2,1,-1,-2,-2,-1,1,2};
    int main()
    {
    scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
    a[x2][y2]=1;
    for(int i=0;i<=7;i++){
    if(x2+mx[i]<=x1&&y2+my[i]<=y1)
    a[x2+mx[i]][y2+my[i]]=1;
    }
    f[0][0]=1;
    for(int i=0;i<=x1;i++)
    for(int j=0;j<=y1;j++){
    if(i>0&&!a[i-1][j])
    f[i][j]+=f[i-1][j];
    if(j>0&&!a[i][j-1])
    f[i][j]+=f[i][j-1];
    }
    printf("%d",f[x1][y1]);
    }
    秒杀。。。。

  • 0
    @ 2015-08-24 16:03:07

    记忆搜索DP
    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <cmath>
    using namespace std;
    typedef long long LL;
    const int N = 300;
    const int M = 10;
    LL dp[N][N];
    int a,b,c,d;
    bool jd(int x,int y)
    {
    if(x == c && y == d)
    return true;
    if(x == (c - 1)&& (y == d + 2|| y == d - 2))
    return true;
    if(x == (c + 1)&& (y == d + 2|| y == d - 2))
    return true;
    if(y == (d + 1) && (x == c + 2 || x == c - 2))
    return true;
    if(y == (d - 1) && (x == c + 2 || x == c - 2))
    return true;
    return false;
    }
    LL DP(int x,int y)
    {
    if( x > a || y > b || jd(x,y))
    return 0;
    if(dp[x][y] != -1)
    return dp[x][y];
    return dp[x][y] = DP(x + 1,y) + DP(x,y+1);
    }
    int main()
    {
    while(~scanf("%d%d%d%d",&a,&b,&c,&d))
    {
    memset(dp,-1,sizeof(dp));
    dp[a][b] = 1;
    printf("%I64d\n",DP(0,0));
    }
    }

  • 0
    @ 2015-08-23 13:24:20

    #include <iostream>
    //#include <fstream>
    using namespace std;
    bool g[30][30];
    long f(int i,int j)
    {
    if(g[i][j])return 0;
    if((!i)||(!j))return 1;
    if(i&&j)return f(i-1,j)+f(i,j-1);
    return 0;
    }
    int main()
    {
    //ifstream cin("p1121.in",ios::in);
    //ofstream cout("p1121.out",ios::out);
    int n,m,x,y;
    cin>>n>>m>>x>>y;
    for(int i=0;i<n;i++)
    for(int j=0;j<m;j++)
    g[i][j]=false;
    g[x][y]=true;
    if((x-2>=0 )&&(x-2<=n )&&(y-1>=0 )&&(y-1<=m)) g[x-2][y-1] =true;
    if((x-2>=0 )&&(x-2<=n )&&(y+1>=0)&&(y+1<=m))g[x-2][y+1] =true;
    if((x-1>=0 )&&(x-1<=n )&&(y+2>=0)&&(y+2<=m))g[x-1][y+2] =true;
    if((x-1>=0 )&&(x-1<=n )&&(y-2>=0 )&&(y-2<=m)) g[x-1][y-2] =true;
    if((x+1>=0)&&(x+1<=n)&&(y+2>=0)&&(y+2<=m))g[x+1][y+2] =true;
    if((x+1>=0)&&(x+1<=n)&&(y-2>=0 )&&(y-2<=m)) g[x+1][y-2] =true;
    if((x+2>=0)&&(x+2<=n)&&(y+1>=0)&&(y+1<=m))g[x+2][y+1] =true;
    if((x+2>=0)&&(x+2<=n)&&(y-1>=0 )&&(y-1<=m)) g[x+2][y-1] =true;
    if((!m)||(!n)||(g[1][0]&&g[0][1]))cout<<0;
    else cout<<f(n,m);
    //cin.close();
    //cout.close();
    return 0;
    }

  • 0
    @ 2015-08-17 17:31:53

    #include<cstdio>
    #include<cstring>
    using namespace std;
    long long g[21][21],f[21][21];
    int main()
    {
    //freopen("knight.in","r",stdin);
    //freopen("knight.out","w",stdout);
    int m,n,i,j,k,a,b;
    scanf("%d%d%d%d",&m,&n,&a,&b);
    memset(g,-1,sizeof(g));
    g[a][b]=0;
    if(a-2>=0&&b-1>=0) g[a-2][b-1]=0;
    if(a-1>=0&&b-2>=0) g[a-1][b-2]=0;
    if(a+1<=m&&b-2>=0) g[a+1][b-2]=0;
    if(a+2<=m&&b-1>=0) g[a+2][b-1]=0;
    if(a-2>=0&&b+1<=n) g[a-2][b+1]=0;
    if(a-1>=0&&b+2<=n) g[a-1][b+2]=0;
    if(a+1<=m&&b+2<=n) g[a+1][b+2]=0;
    if(a+2<=m&&b+1<=n) g[a+2][b+1]=0;
    memset(f,0,sizeof(f));
    f[1][0]=1;
    f[0][1]=1;
    if(g[1][0]==0){
    f[1][0]=0;
    }
    if(g[0][1]==0){
    f[0][1]=0;
    }
    for(i=2;i<=m+n;i++)
    {
    j=0;
    k=i;
    while(j<=i&&k>=0)
    {
    if(j<=m&&k<=n&&g[j][k]!=0)
    {
    if(j-1>=0)
    f[j][k]+=f[j-1][k];
    if(k-1>=0)
    f[j][k]+=f[j][k-1];
    /*printf("f[%d][%d]=",j,k);
    printf("%I64d\n",f[j][k]);*/
    }
    j++;
    k--;
    }
    }
    printf("%I64d\n",f[m][n]);
    return 0;
    }

  • 0
    @ 2015-08-17 10:35:14

    #include <iostream>
    #include <algorithm>
    #include <cstdio>
    using namespace std;

    int f[25][25],i,j,n,m,x,y,x1,x2,l;
    bool a[25][25];

    int dp(int x2,int y2)
    {
    if (x2==x && y2==y)
    {
    l++;
    return 0;
    }
    if (x2<x && a[x2+1][y2]==true)
    {

    dp(x2+1,y2);
    }
    if (y2<y && a[x2][y2+1]==true)
    {

    dp(x2,y2+1);
    }
    if (y2>y && a[x2][y2-1]==true)
    {

    dp(x2,y2-1);
    }
    }

    int main()
    {
    cin>>x>>y>>x1>>x2;
    for (i=0;i<=15;i++)
    for (j=0;j<=15;j++)
    a[i][j]=true;
    a[x1][x2]=false;
    if (x2-1>=0) a[x1+2][x2-1]=false;
    a[x1+2][x2+1]=false;
    if (x2-1>=0 && x1-2>=0) a[x1-2][x2-1]=false;
    if (x1-2>=0) a[x1-2][x2+1]=false;
    a[x1+1][x2+2]=false;
    if (x1-1>=0) a[x1-1][x2+2]=false;
    if (x2-2>=0 && x1-1>=0) a[x1-1][x2-2]=false;
    if (x2-2>=0) a[x1+1][x2-2]=false;
    dp(0,0);
    cout <<l;
    }

    • @ 2015-08-17 10:36:07

      我只是把过程名dfs改成了dp而已 O(∩_∩)O

    • @ 2015-10-03 16:16:32

      这道题只不过是抄我的而已。。。。

  • 0
    @ 2015-08-17 09:24:20

    有什么问题就找那个用户名叫黎明前的光的大神

  • 0
    @ 2015-08-16 16:49:17

    #include<iostream>
    #include<cstring>
    using namespace std;

    const int maxn=20;
    int d[maxn][maxn],n,m,x,y,vis[maxn][maxn]={0};

    bool pd(int i,int j) {
    if (i==x-2&&j==y-1) return 1;
    if (i==x-2&&j==y+1) return 1;
    if (i==x+2&&j==y-1) return 1;
    if (i==x+2&&j==y+1) return 1;
    if (i==x-1&&j==y-2) return 1;
    if (i==x-1&&j==y+2) return 1;
    if (i==x+1&&j==y-2) return 1;
    if (i==x+1&&j==y+2) return 1;
    if (i==x&&j==y) return 1;
    return 0;
    }

    int dp(int i,int j) {
    if (i>n||j>m) return -1;
    else if (pd(i,j)) return -1;
    else if (i==n&&j==m) return d[i][j]=1;
    else if (vis[i][j]) return d[i][j];
    int &ans=d[i][j];
    vis[i][j]=1;
    if (dp(i+1,j)!=-1) {
    if (ans==-1) ans=0;
    ans+=d[i+1][j];
    }
    if (dp(i,j+1)!=-1) {
    if (ans==-1) ans=0;
    ans+=d[i][j+1];
    }
    // cout<<"d["<<i<<','<<j<<"]="<<d[i][j]<<endl;
    return ans;
    }

    int main() {
    memset(d,-1,sizeof(d));
    cin>>n>>m>>x>>y;
    dp(0,0);
    cout<<(d[0][0]==-1?0:d[0][0]);
    return 0;
    }

  • 0
    @ 2015-08-09 16:33:30

    #include <iostream>
    #include <stdio.h>
    #include <string.h>

    using namespace std;

    int n , m;
    int a , b;

    int f[15 + 5][15 + 5];

    void mark( int a , int b )
    {
    if( a >= 0 && b >= 0 )
    f[a][b] = 0;
    return;
    }

    int dp( int a , int b )
    {
    if( a < 0 || b < 0 )
    return 0;
    if( f[a][b] != -1 )
    return f[a][b];
    if( a == 0 && b == 0 )
    return 1;
    return f[a][b] = dp( a - 1 , b ) + dp( a , b - 1 );
    }

    int main()
    {
    scanf( "%d %d" , &n , &m );
    scanf( "%d %d" , &a , &b );
    memset( f , -1 , sizeof( f ) );
    mark( a - 2 , b - 1 );
    mark( a - 2 , b + 1 );
    mark( a + 2 , b - 1 );
    mark( a + 2 , b + 1 );
    mark( a - 1 , b - 2 );
    mark( a - 1 , b + 2 );
    mark( a + 1 , b - 2 );
    mark( a + 1 , b + 2 );
    mark( a , b );
    cout << dp( n , m ) << endl;
    return 0;
    }
    小时候一直没有做对的题

  • 0
    @ 2015-06-29 11:16:54

    #include<cstdio>
    #include<iostream>
    #include<cstring>
    using namespace std;
    int s=0,n,m;
    int a[50][50];
    void dfs(int x,int y){
    if ((x==n)&&(y==m)) s++;
    if ((x+1<=n)&&(a[x+1][y]==0)) {
    a[x+1][y]=1;
    dfs(x+1,y);
    a[x+1][y]=0;
    }
    if ((y+1<=m)&&(a[x][y+1]==0)) {
    a[x][y+1]=1;
    dfs(x,y+1);
    a[x][y+1]=0;
    }
    }
    int main()
    {
    int x,y;
    scanf("%d %d %d %d",&n,&m,&x,&y);
    memset(a,0,sizeof(a));
    a[x][y]=1;
    a[x-2][y-1]=1;
    a[x-1][y-2]=1;
    a[x-2][y+1]=1;
    a[x-1][y+2]=1;
    a[x+2][y-1]=1;
    a[x+1][y-2]=1;
    a[x+2][y+1]=1;
    a[x+1][y+2]=1;
    dfs(0,0);
    printf("%d",s);
    return 0;
    }

  • 0
    @ 2015-06-24 22:19:18

    C++ Code

    #include <iostream>
    using namespace std;

    int v[16][16];

    int max(int x1,int x2)
    {
    return x1>x2?x1:x2;
    }

    int hx,hy,x,y;
    bool ava(int x1,int y1)
    {
    if
    (
    (x1==hx && y1==hy)
    ||
    ((x1+1==hx||x1-1==hx)&&(y1+2==hy||y1-2==hy))
    ||
    ((y1+1==hy||y1-1==hy)&&(x1+2==hx||x1-2==hx))
    )
    {
    return false;
    }
    return true;
    }
    int main()
    {
    cin>>x>>y>>hx>>hy;
    hx++;
    hy++;
    x++;
    y++;

    if(ava(1,1))v[1][1]=1;
    for(int i=2;i<=x;i++)
    {
    v[i][1]=v[i-1][1];
    if(ava(i,1)==false)v[i][1]=0;
    }
    for(int i=2;i<=y;i++)
    {
    v[1][i]=v[1][i-1];
    if(ava(1,i)==false)v[1][i]=0;
    }
    for(int xx=2;xx<=x;xx++)
    {
    for(int yy=2;yy<=y;yy++)
    {
    if(ava(xx,yy))
    {
    v[xx][yy]=v[xx-1][yy]+v[xx][yy-1];
    }
    else v[xx][yy]=0;
    }
    }

    cout<<v[x][y]<<endl;
    return 0;
    }

  • 0
    @ 2015-06-12 21:44:51

    非常水呀

  • 0
    @ 2015-06-12 21:44:15

    program p1287;
    var
    a:array[-1..21,-1..21] of longint;
    i,j,k,m,n,x,y:longint;
    begin
    readln(n,m,x,y);
    a[x,y]:=-1;
    a[x+1,y+2]:=-1;
    a[x+1,y-2]:=-1;
    a[x+2,y-1]:=-1;
    a[x+2,y+1]:=-1;
    a[x-1,y-2]:=-1;
    a[x-1,y+2]:=-1;
    a[x-2,y-1]:=-1;
    a[x-2,y+1]:=-1;
    a[0,0]:=1;
    for i:=0 to n do
    for j:=0 to m do
    begin
    if a[i,j]=0 then
    begin
    a[i,j]:=a[i-1,j]+a[i,j-1];
    if a[i-1,j]=-1 then inc(a[i,j]);
    if a[i,j-1]=-1 then inc(a[i,j]);
    end;
    end;
    writeln(a[n,m]);
    end.

信息

ID
1121
难度
4
分类
动态规划 点击显示
标签
递交数
9572
已通过
3779
通过率
39%
被复制
23
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