Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 673 Accepted Submission(s): 223
int sum(int x,int y){
int ret=0;
for(int i=x;i>0;i-=lowbit(i)){
for(int j=y;j>0;j-=lowbit(j)){
ret+=C[i][j];
}
}
return ret;
}
修改:
void modify(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;
}
}
}

#include<bits/stdc++.h>
using namespace std;
typedef long long INT;
const int maxn=550;
int a[maxn][maxn];
int C[maxn][maxn];
int n,m;
int lowbit(int x){
return x & (-x);
}
void modify(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;
}
}
}
int sum(int x,int y){
int ret=0;
for(int i=x;i>0;i-=lowbit(i)){
for(int j=y;j>0;j-=lowbit(j)){
ret ^=C[i][j];
}
}
return ret;
}
int main(){
int t,q;
scanf("%d",&t);
while(t--){
memset(a,0,sizeof(a));
memset(C,0,sizeof(C));
scanf("%d%d%d",&n,&m,&q);
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
scanf("%d",&a[i][j]);
}
}
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
modify(i,j,a[i][j]);
}
}
int typ,x1,x2,y1,y2,c;
for(int i=0;i<q;i++){
scanf("%d",&typ);
if(typ==1){
scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
int ans1,ans2,ans3,ans4;
ans1=sum(x2,y2);
ans2=sum(x2,y1-1);
ans3=sum(x1-1,y2);
ans4=sum(x1-1,y1-1);
int ans=ans1^ans2^ans3^ans4;
printf("%s\n",ans==0?"No":"Yes");
}else{
scanf("%d%d%d",&x1,&y1,&c);
modify(x1,y1,c^a[x1][y1]);
a[x1][y1]=c;
}
}
}
return 0;
}
HDU 5465——Clarke and puzzle——————【树状数组BIT维护前缀和+Nim博弈】
原文:http://www.cnblogs.com/chengsheng/p/4832894.html