解析：
利用二维树状数组来区间询问异或和，以及单点更新，然后利用nim博弈的结论判断胜负。
my code
#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef long long ll;const int N = 505;int n, m, q;struct BIT {int c[N][N];void clear() { memset(c, 0, sizeof(c)); }inline int lowbit(int x) { return x & (-x); }int query(int x, int y) {int res = 0;for(int i = x; i > 0; i -= lowbit(i)) {for(int j = y; j > 0; j -= lowbit(j)) {res ^= c[i][j];}}return res;}int query(int x1, int y1, int x2, int y2) {x1--, y1--;int area[2][2] = {0};area[0][0] = query(x1, y1); area[1][1] = query(x2, y2);area[0][1] = query(x1, y2); area[1][0] = query(x2, y1);int res = 0;for(int i = 0; i < 2; i++)for(int j = 0; j < 2; j++)res ^= area[i][j];return res;}void add(int x, int y, int val) {for(int i = x; i <= n; i += lowbit(i))for(int j = y; j <= m; j += lowbit(j))c[i][j] ^= val;}void modify(int x, int y, int val) {add(x, y, query(x, y, x, y));add(x, y, val);}} bit;int main() {int op, x1, x2, y1, y2, val;int T;scanf("%d", &T);while(T--) {scanf("%d%d%d", &n, &m, &q);bit.clear();for(int i = 1; i <= n; i++) {for(int j = 1; j <= m; j++) {scanf("%d", &val);bit.add(i, j, val);}}while(q--) {scanf("%d", &op);if(op == 1) {scanf("%d%d%d%d", &x1, &y1, &x2, &y2);int ans = bit.query(x1, y1, x2, y2);puts(ans != 0 ? "Yes" : "No");}else {scanf("%d%d%d", &x1, &y1, &val);bit.modify(x1, y1, val);}}}return 0;}
