二维树状数组模板(区间修改+区间查询)
例题：JOIOI上帝造题的七分钟
一共两种操作：
\(L\ x_1\ y_1\ x_2\ y_2\ d\)：把\((x_1,y_1)\)，\((x_2,y_2)\)这个矩形内所有元素加\(d\)。
\(k\ x_1\ y_1\ x_2\ y_2\)：查询\((x_1,y_1)\)，\((x_2,y_2)\)这个矩形内所有元素的和。
代码如下：
#include
#define RG register
#define IL inline
#define _ 2050
#define ll long long
//#define ll int
using namespace std;
ll n , m , c1[_][_],c2[_][_],c3[_][_],c4[_][_] ;
IL void upt(ll x , ll y , ll dt){
for(RG int i = x; i <= n; i += (i & -i) )
for(RG ll j = y; j <= m; j += (j & -j)){
c1[ i ][ j ] += dt ;
c2[ i ][ j ] += dt * y;
c3[ i ][ j ] += dt * x;
c4[ i ][ j ] += dt * x * y;
}
}
IL ll calc(ll x,ll y){
RG ll res = 0;
for(RG ll i = x; i > 0; i -= (i & -i))
for(RG ll j = y; j > 0; j -= (j & -j)){
res = res
+ (x + 1) * (y + 1) * c1[ i ][ j ]
- (x + 1) * c2[ i ][ j ]
- (y + 1) * c3[ i ][ j ]
+ c4[ i ][ j ] ;
}
return res;
}
IL void add(ll X1,ll Y1,ll X2,ll Y2,ll dt){
upt(X1 , Y1 , dt ) ;
upt(X2 + 1 , Y1 , -dt ) ;
upt(X1 , Y2 + 1, -dt ) ;
upt(X2 + 1, Y2 + 1, dt ) ;
}
IL ll query(ll X1,ll Y1,ll X2,ll Y2){
return
calc(X2 , Y2) + calc(X1 - 1, Y1 - 1) -
calc(X1 - 1 , Y2) - calc(X2 , Y1 - 1) ;
}
int main(){
ll X1 , X2 , Y1 , Y2 , z; char c[3];
scanf("X %lld %lld",&n,&m);
while(scanf("%s",c)!=EOF){
scanf("%lld%lld%lld%lld", &X1, &Y1, &X2, &Y2);
if(c[0]=='L'){
scanf("%lld", &z);
add(X1 , Y1 , X2 , Y2 , z);
}
else printf("%lld\n",query(X1 , Y1 , X2 , Y2 ));
}return 0;return 0;
}