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\n", 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("\n");
    }

    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("\n");
//    }

    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(">>>>>>\n");
    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("\n");
    }
}

　　

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]);
        }
    }