# include <bits/stdc++.h>
# define int long long
using namespace std;
const int TT=10000;
int n,m;
int random(int x) { return rand()*rand()%x;}
int Fun(int low,int high)
{ return random(10000000) % (high - low + 1) + low;} 
void work1()
{
    int x=random(n)+1,y=random(m)+1;
    printf("F %d %d\n",x,y); 
}
void work2()
{
    int x1=random(n)+1,y1=random(m)+1;
    int x2=Fun(x1,n),y2=Fun(y1,m);
    printf("Q %d %d %d %d\n",x1,y1,x2,y2);
}
void work3()
{
    int x1=random(n)+1,y1=random(m)+1;
    int x2=Fun(x1,n),y2=Fun(y1,m);
    int d=random(2*TT+1)-TT;
    printf("A %d %d %d %d %d\n",x1,y1,x2,y2,d);
}
void work4()
{
    int x1=random(n)+1,y1=random(m)+1;
    int x2=Fun(x1,n),y2=Fun(y1,m);
    int d=random(2*TT+1)-TT;
    printf("D %d %d %d %d %d\n",x1,y1,x2,y2,d);
}
signed main()
{
    srand((unsigned long long)time(NULL)*19260817);
    puts("宽长："); scanf("%d%d",&n,&m);
    int F,Q,A,D,R,T;
    puts("输入操作个数："); scanf("%d",&T);
    puts("输入F操作的几率:");scanf("%d",&F);
    puts("输入Q操作的几率:");scanf("%d",&Q);
    puts("输入A操作的几率:");scanf("%d",&A);
    puts("输入D操作的几率:");scanf("%d",&D); 
    puts("输入干扰操作个数:");scanf("%d",&R);
    freopen("7.in","w",stdout);
    printf("%d %d\n",n,m);
    for (int i=1;i<=n;i++) {
      for (int j=1;j<=m;j++)
       printf("%d ",random(2*TT+1)-TT);
       puts("");
    }
    while (T--) {
        int opt=random(100);
        if (opt<F) work1();
        else if (opt<F+Q) work2();
        else if (opt<F+Q+A) work3();
        else work4();
    }
    puts("E");
    while (R--) {
        int opt=random(100);
        if (opt<F) work1();
        else if (opt<F+Q) work2();
        else if (opt<F+Q+A) work3();
        else work4();
    } 
    return 0;
}

solution:
#5 5%的数据 打卡数据
#1-#4 20%数据
不采用快速读入输出的模拟
#11-#12 10%数据
采用快速读入输出的模拟(不就是卡了常数嘛)
暴力程序：

# include <bits/stdc++.h>
# define int long long 
using namespace std;
const int MAXN=5005;
int a[MAXN][MAXN],n,m;
inline int read()
{
    int X=0,w=0;char c=0;
    while (!(c>='0'&&c<='9')) w|=c=='-',c=getchar();
    while ((c>='0'&&c<='9')) X=(X<<1)+(X<<3)+(c^48),c=getchar();
    return w?-X:X;
}
inline void write(int x)
{
    if (x<0) { putchar('-'); x=-x;}
    if (x>9) write(x/10);
    putchar('0'+x%10);
}
inline void writeln(int x){write(x);putchar('\n');}
inline void work1()
{
    int ans=0;
    int x1,y1,x2,y2;
    x1=read(),y1=read(),x2=read(),y2=read();
    for (int i=x1;i<=x2;i++)
     for (int j=y1;j<=y2;j++)
      ans+=a[i][j];
    writeln(ans);
}
inline void work2()
{
    int x1,y1,x2,y2,d;
    x1=read(),y1=read(),x2=read(),y2=read(),d=read();
    for (int i=x1;i<=x2;i++)
     for (int j=y1;j<=y2;j++)
      a[i][j]+=d;
}
inline void work3()
{
    int x1,y1,x2,y2,d;
    x1=read(),y1=read(),x2=read(),y2=read(),d=read();
    for (int i=x1;i<=x2;i++)
     for (int j=y1;j<=y2;j++)
      a[i][j]-=d;
}
inline void work4()
{
    int x,y;
    x=read();y=read();
    writeln(a[x][y]);
}
signed main()
{
    n=read();m=read();
    for (int i=1;i<=n;i++)
     for (int j=1;j<=m;j++)
      a[i][j]=read();
    char opt;  
    while (true) {
        cin>>opt;
        switch (opt) {
            case 'Q':work1();break;
            case 'A':work2();break;
            case 'D':work3();break;
            case 'F':work4();break;
            case 'E':goto exit;break;
        }
    }  
    exit:;
    return 0;
}

#6-#7 10%的数据维护区间加区间求和考虑线段树

#8-#10 15%的数据没有区间求和操作，维护一颗二维树状数组

#13-#20 40%的数据正解

# pragma G++ optimze(2)
# include<bits/stdc++.h>
# define int long long
using namespace std;
const int MAXN=5005;
int a[MAXN][MAXN],d[MAXN][MAXN];
int n,m;
inline int read()
{
    int X=0,w=0;char c=0;
    while (!(c>='0'&&c<='9')) w|=c=='-',c=getchar();
    while ((c>='0'&&c<='9')) X=(X<<1)+(X<<3)+(c^48),c=getchar();
    return w?-X:X;
}
inline void write(int x)
{
    if (x<0) { putchar('-'); x=-x;}
    if (x>9) write(x/10);
    putchar('0'+x%10);
}
inline void writeln(int x){write(x);putchar('\n');}
struct Tree{
    int c[MAXN][MAXN];
    Tree(){ memset(c,0,sizeof(c));}
    inline void update(int x,int y,int z){
        for (int i=x;i<=n;i+=i&(-i))
         for (int j=y;j<=m;j+=j&(-j)){
            c[i][j]+=z;
         }
    }   
}T1,T2,T3,T4;
# define A (x+1)
# define B (y+1)
inline int ask(int x,int y){
    int ret=0;
    for (int i=x;i;i-=i&(-i))
     for (int j=y;j;j-=j&(-j))
      ret+=A*B*T1.c[i][j]-B*T2.c[i][j]-A*T3.c[i][j]+T4.c[i][j];
    return ret;  
}
inline void add(int x,int y,int val){
    T1.update(x,y,val);   T2.update(x,y,val*x);
    T3.update(x,y,val*y); T4.update(x,y,val*x*y);
}
inline void work1()
{
    int x1=read(),y1=read(),x2=read(),y2=read();
    int ans=ask(x2,y2)-ask(x2,y1-1)-ask(x1-1,y2)+ask(x1-1,y1-1);
    writeln(ans);
}
inline void work2()
{
    int x1=read(),y1=read(),x2=read(),y2=read(),d=read();
    add(x1,y1,d); add(x1,y2+1,-d);
    add(x2+1,y1,-d); add(x2+1,y2+1,d);
}
inline void work3()
{
    int x1=read(),y1=read(),x2=read(),y2=read(),d=read(); d=-d;
    add(x1,y1,d); add(x1,y2+1,-d);
    add(x2+1,y1,-d); add(x2+1,y2+1,d);
}
inline void work4()
{
    int x1=read(),y1=read();
    int x2=x1,y2=y1;
    int ans=ask(x2,y2)-ask(x2,y1-1)-ask(x1-1,y2)+ask(x1-1,y1-1);
    writeln(ans);
}
signed main()
{
    #ifdef LOCAL
        freopen("input.in","r",stdin);
        freopen("output.out","w",stdout);
    #endif
    n=read();m=read();
    for (int i=1;i<=n;i++)
     for (int j=1;j<=m;j++) {
        a[i][j]=read();
        d[i][j]=a[i][j]-a[i-1][j]-a[i][j-1]+a[i-1][j-1];  
     }
     for (int i=1;i<=n;i++)
      for (int j=1;j<=m;j++)
        add(i,j,d[i][j]);
      char opt;
      while (true) {
        cin>>opt;
        switch (opt) {
            case 'Q':work1();break;
            case 'A':work2();break;
            case 'D':work3();break;
            case 'F':work4();break;
            case 'E':goto exit;break;   
        }
      }
      exit:;
    return 0;
}