题意：一个n*m初始为空的矩阵，资磁两种操作：（1）某个子矩阵的值都增加c （2）输出某个子矩阵的权值和
用a[i,j]表示（i,j）~（n,m）的增量，则对于（1,1）~（x,y）的权值和ans
ans=sigma（a[i,j]*(x-i+1）*（y-j+1））(1<=i<=x,1<=j<=y)
=sigma(a[i,j]*(x+1)*(y+1)-j*(x+1)*a[i,j]-i*(y+1)*a[i,j]+i*j*a[i,j]) (1<=i<=x,1<=j<=y)
=(x+1)*(y+1)*sigma(a[i,j])-(x+1)*sigma(j*a[i,j])-(y+1)*sigma(i*a[i,j])+sigma(i*j*a[i,j]) (1<=i<=x,1<=j<=y)
维护四个树状数组分别为
1：sigma(a[i,j])
2：sigma(j*a[i,j])
3：sigma(i*a[i,j])
4：sigma(i*j*a[i,j])
{$Q-} var n,m,x1,x2,y1,y2,c :longint; ch :char; t :array[0..4,0..2105,0..2105] of longint; procedure _add(z,x,y,v:longint); var tt:longint; begin while x<=n do begin tt:=y; while tt<=m do begin inc(t[z,x,tt],v); inc(tt,tt and (-tt)); end; inc(x,x and (-x)); end; end; procedure change(x1,y1,x2,y2,c:longint); begin _add(1,x1,y1,c); _add(1,x2+1,y2+1,c); _add(1,x1,y2+1,-c); _add(1,x2+1,y1,-c); // _add(2,x1,y1,y1*c); _add(3,x1,y1,x1*c); _add(2,x2+1,y2+1,(y2+1)*c); _add(3,x2+1,y2+1,(x2+1)*c); _add(2,x1,y2+1,-(y2+1)*c); _add(3,x1,y2+1,-x1*c); _add(2,x2+1,y1,-y1*c); _add(3,x2+1,y1,-(x2+1)*c); // _add(4,x1,y1,x1*y1*c); _add(4,x2+1,y2+1,(x2+1)*(y2+1)*c); _add(4,x1,y2+1,-x1*(y2+1)*c); _add(4,x2+1,y1,-(x2+1)*y1*c); end; function find(z,x,y:longint):longint; var tt,ans:longint; begin ans:=0; while x>0 do begin tt:=y; while tt>0 do begin inc(ans,t[z,x,tt]); dec(tt,tt and (-tt)); end; dec(x,x and (-x)); end; exit(ans); end; function sum(x,y:longint):longint; begin exit(find(1,x,y)*(x+1)*(y+1)-(x+1)*find(2,x,y)-(y+1)*find(3,x,y)+find(4,x,y)); end; begin read(ch); readln(n,m); while not eof do begin read(ch); if ch='L' then begin readln(x1,y1,x2,y2,c); change(x1,y1,x2,y2,c); end else begin readln(x1,y1,x2,y2); writeln(sum(x2,y2)-sum(x1-1,y2)-sum(x2,y1-1)+sum(x1-1,y1-1)); end; end; end. ——by Eirlys
