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  • 二维数组旋转45度

    //#pragma comment(linker, "/STACK:1024000000,1024000000")
    #include <iostream>
    #include <cstdio>
    #include <cmath>
    #include <vector>
    #include <cstring>
    #include <algorithm>
    #include <string>
    #include <set>
    #include <functional>
    #include <numeric>
    #include <sstream>
    #include <stack>
    #include <map>
    #include <queue>
    
    #define CL(arr, val)    memset(arr, val, sizeof(arr))
    
    #define lc l,m,rt<<1
    #define rc m + 1,r,rt<<1|1
    #define pi acos(-1.0)
    #define ll long long
    #define L(x)    (x) << 1
    #define R(x)    (x) << 1 | 1
    #define MID(l, r)   (l + r) >> 1
    #define Min(x, y)   (x) < (y) ? (x) : (y)
    #define Max(x, y)   (x) < (y) ? (y) : (x)
    #define E(x)        (1 << (x))
    #define iabs(x)     (x) < 0 ? -(x) : (x)
    #define OUT(x)  printf("%I64d
    ", x)
    #define keyTree (chd[chd[root][1]][0])
    #define Read()  freopen("din.txt", "r", stdin)
    #define Write() freopen("dout.txt", "w", stdout);
    
    
    #define M 100007
    #define N 107
    
    using namespace std;
    
    int dx[4]={-1,1,0,0};
    int dy[4]={0,0,-1,1};
    
    const int inf = 0x7f7f7f7f;
    const int mod = 1000000007;
    const double eps = 1e-8;
    const int R = 100007;
    
    int a[N][N];
    vector<int> b[2*N];
    
    int main()
    {
        int cnt = 0;
        int n ;
        cin >> n;
        for (int i = 0; i < 2*n - 1; ++i) b[i].clear();
    
        for (int i = 0; i < n; ++i)
        {
            for (int j = 0; j < n; ++j)
            {
                a[i][j] = cnt++;
            }
        }
        for (int i = 0; i < n; ++i)
        {
            for (int j = 0; j < n; ++j)
            {
                printf("%d ",a[i][j]);
            }
            printf("
    ");
        }
    
        for (int i = 0; i < n; ++i)
        {
            for (int j = 0; j <= i; ++j)
            {
                //b[i][j] = a[j][i - j];
                b[i].push_back(a[j][i - j]);
            }
        }
        //输出三角的旋转
    //    for (int i = 0; i < n; ++i)
    //    {
    //        for (int j = 0; j <= i; ++j)
    //        {
    //            printf("%d ",b[i][j]);
    //        }
    //        printf("
    ");
    //    }
    
        for (int i = n,rnum = n - 1; i < 2*n - 1; ++i,rnum--)
        {
            for (int j = 0; j < rnum; j++)
            {
                //b[i][j] = a[i - n + 1 + j][n - 1 - j];
                b[i].push_back(a[i - n + 1 + j][n - 1 - j]);
            }
        }
        printf(">>>>>>
    ");
        for (int i = 0; i < 2*n - 1; ++i)
        {
            for (size_t j = 0; j < b[i].size(); ++j)
            {
                printf("%02d ",b[i][j]);
            }
            printf("
    ");
        }
    }
    

      

    还是这个比较简单:

    for (int i = 0; i < 2*n - 1; ++i)
        {
            for (int j = 0; j < n; ++j)
            {
                if (i - j < 0 || i - j >= n) continue;
                b[i].push_back(a[j][i - j]);
            }
        }
    

      

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  • 原文地址:https://www.cnblogs.com/E-star/p/3369773.html
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