Codeforces 669C Little Artem and Matrix (模拟)

题意

对一个矩阵有三种操作
1.第row行循环右移x位
2.第col列循环下移x位
3.告诉你第row行的第col列的数是x。
要求构造出符合操作的原矩阵。

思路

因为数据很小，直接暴力就可以了，按照操作倒着模拟一遍，有3操作的就填数，没有的就为0。

代码

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <stack>#include <set>#include <map>#include <string>#include <math.h>#include <stdlib.h>#include <time.h>using namespace std;#define LL long long#define Lowbit(x) ((x)&(-x))#define lson l, mid, rt << 1#define rson mid + 1, r, rt << 1|1#define MP(a, b) make_pair(a, b)const int INF = 0x3f3f3f3f;const int MOD = 1000000007;const int maxn = 1e5 + 10;const double eps = 1e-8;const double PI = acos(-1.0);typedef pair<int, int> pii;int a[110][110];struct node{    int type;    int r, c, x;}p[10010];int main(){    //freopen("in.txt","r",stdin);    //freopen("out.txt","w",stdout);    memset(a, 0, sizeof(a));    int n, m, q;    scanf("%d%d%d", &n, &m, &q);    for (int i = 0; i < q; i++)    {        int t;        scanf("%d", &t);        if (t == 3)        {            p[i].type = 3;            scanf("%d%d%d", &p[i].r, &p[i].c, &p[i].x);        }        if (t == 2)        {            p[i].type = 2;            scanf("%d", &p[i].c);        }        if (t == 1)        {            p[i].type = 1;            scanf("%d", &p[i].r);        }    }    for (int i = q - 1; i >= 0; i--)    {        if (p[i].type == 1)        {            int row = p[i].r;            int t = a[row][m];            for (int i = m; i >= 2; i--)                a[row][i] = a[row][i-1];            a[row][1] = t;        }        if (p[i].type == 2)        {            int col = p[i].c;            int t = a[n][col];            for (int i = n; i >= 2; i--)                a[i][col] = a[i-1][col];            a[1][col] = t;        }        if (p[i].type == 3)        {            a[p[i].r][p[i].c] = p[i].x;        }    }    for (int i = 1; i <= n; i++, printf("\n"))        for (int j = 1; j <= m; j++)            printf("%d ", a[i][j]);    return 0;}