两个操作:
1. Add x y value: Add value to the element Axy. (Subscripts starts from 0
2. Sum x1 y1 x2 y2: Return the sum of every element Axy for x1 ≤ x ≤ x2, y1 ≤ y ≤ y2.
注意取模,因为value可能为负值
ans%=Mod;
if(ans<0) ans+=Mod;
#include <iostream>
#include <cstdio>
using namespace std;
const int Max = 1010;
const int Mod = 1e9+7;
typedef long long LL;
int n;
int c[1010][1010];
int lowbit(int x)
{
return x&(-x);
}
void update(int x,int y,int val)
{
for(int i=x;i<=Max;i+=lowbit(i))
{
for(int j=y;j<=Max;j+=lowbit(j))
{
c[i][j]+=val;
}
}
}
LL sum(int x,int y)
{
LL 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 main()
{
int op,x,y,X,Y,val;
string str;
scanf("%d%d",&n,&op);
while(op--)
{
cin>>str;
if(str[0]=='A'){
scanf("%d%d%d",&x,&y,&val);
x++,y++;
update(x,y,val);
}else{//sum
scanf("%d%d%d%d",&x,&y,&X,&Y);
x++,y++,X++,Y++;
LL ans =sum(X,Y)+sum(x-1,y-1)-sum(X,y-1)-sum(x-1,Y);
ans%=Mod;
if(ans<0) ans+=Mod;
printf("%d
",ans);
}
}
return 0;
}