二维数组前缀和

对于二维数组求前缀和,我们先预处理第一行跟第一列的前缀和

第一行跟第一列的前缀和可以看作一维数组的前缀和

前缀和数组的0,0等于原数组的0,0,第一行为原数组第一行的前缀和,第一列为第一列的前缀和

预处理:

//原数组
int arr[110][110]={{},{0,1,2,3,4,5,6,7,8,9,0},{0,6,7,8,9,0,1,2,3,4,5},{0,1,2,3,4,5,6,7,8,9,0},{0,1,2,3,4,5,6,7,8,9,0},{0,1,2,3,4,5,6,7,8,9,0},{0,1,2,3,4,5,6,7,8,9,0},{0,1,2,3,4,5,6,7,8,9,0},{0,1,2,3,4,5,6,7,8,9,0},{0,1,2,3,4,5,6,7,8,9,0},{0,1,2,3,4,5,6,7,8,9,0},
};
int sum[110][110]={0};//前缀和数组//预处理
sum[1][1]=arr[1][1];
for(int i=2;i<=10;i++) {sum[1][i]=arr[1][i]+sum[1][i-1];sum[i][1]=arr[i][1]+sum[i-1][1];
}

遍历剩下的部分求前缀和:

如图不难看出$(x,y)$前缀和等于$(x-1,y)$的前缀和加上$(x,y-1)$的前缀和再减去一次重叠部分$(x-1,y-1)$的前缀和,最后再加上数组本身的$(x,y)$

代码:

//原数组
int arr[110][110]={{},{0,1,2,3,4,5,6,7,8,9,0},{0,6,7,8,9,0,1,2,3,4,5},{0,1,2,3,4,5,6,7,8,9,0},{0,1,2,3,4,5,6,7,8,9,0},{0,1,2,3,4,5,6,7,8,9,0},{0,1,2,3,4,5,6,7,8,9,0},{0,1,2,3,4,5,6,7,8,9,0},{0,1,2,3,4,5,6,7,8,9,0},{0,1,2,3,4,5,6,7,8,9,0},{0,1,2,3,4,5,6,7,8,9,0},
};int sum[110][110]={0};//前缀和数组//预处理
sum[1][1]=arr[1][1];
for(int i=2;i<=10;i++) {sum[1][i]=arr[1][i]+sum[1][i-1];sum[i][1]=arr[i][1]+sum[i-1][1];
}//剩下的部分求和代码
for(int i=2;i<=10;i++){for(int o=2;o<=10;o++){sum[i][o]=sum[i-1][o]+sum[i][o-1]-sum[i-1][o-1]+arr[i][o];}
}

这样就求出来二维数组的前缀和

区间和:

求二维数组$(x1,y1)->(x2,y2)$的和$(x1<=x2)(y1<=y2)$

求出前缀和之后,我们可以以$O(1)$的速度求出来

求$(x1,y1)->(x2,y2)$的值应该等于$(x2,y2)$的前缀和减去($x1-1,y2),(x2,y1-1)$的前缀和再加上重叠部分$(x1-1,y1-1)$的和就是$(x1,y1)->(x2,y2)$区间的和

代码:

int n;
cin>>n;
while(n-->0){int x1,y1,x2,y2;cin>>x1>>y1>>x2>>y2;int S=sum[x2][y2]-sum[x1-1][y2]-sum[x2][y1-1]+sum[x1-1][y1-1];cout<<S<<endl;
}

n次询问,我们都能以O(1)的速度求出区间的和