偶然发现这题还没A掉............速速解决了.............
树状数组和线段树比较下，线段树是在是太冗余了，以后能用树状数组还是尽量用.........
#include#include #include #include #include #include #include #include #include #include #include #include //形如INT_MAX一类的#define MAX 1111#define INF 0x7FFFFFFF#define REP(i,s,t) for(int i=(s);i<=(t);++i)#define ll long long#define mem(a,b) memset(a,b,sizeof(a))#define mp(a,b) make_pair(a,b)#define L(x) x<<1#define R(x) x<<1|1# define eps 1e-5//#pragma comment(linker, "/STACK:36777216") ///传说中的外挂using namespace std;int c[MAX][MAX],n;int lowbit(int x) { return x & (-x);}void update(int x1,int y1,int va) { for(int i=x1; i<=n+1; i += lowbit(i)) { for(int j=y1; j<=n+1; j += lowbit(j)) { c[i][j] += va; } }}int query(int x1,int y1) { int sum = 0; for(int i=x1; i>0; i -= lowbit(i)) { for(int j=y1; j>0; j -= lowbit(j)) { sum += c[i][j]; } } return sum;}int main() { int op,x1,y1,x2,y2,va; scanf("%d%d",&op,&n); memset(c,0,sizeof(c)); while(scanf("%d",&op) ) { if(op == 3) break; if(op == 1) { scanf("%d%d%d",&x1,&y1,&va); update(x1+1,y1+1,va); } if(op == 2) { scanf("%d%d%d%d",&x1,&y1,&x2,&y2); printf("%d\n",query(x2+1,y2+1) - query(x1,y2+1) + query(x1,y1) - query(x2+1,y1)); } } return 0;}