杭电2018多校第四场(2018 Multi-University Training Contest 4) 1005.Problem E. Matrix from Arrays (HDU6336) -子矩阵求和-规律+二维前缀和

 1 //1005-6336-矩阵求和-二维前缀和+容斥-预处理O(1)查询输出
 2 #include<iostream>
 3 #include<cstdio>
 4 #include<cstring>
 5 #include<algorithm>
 6 #include<bitset>
 7 #include<cassert>
 8 #include<cctype>
 9 #include<cmath>
10 #include<cstdlib>
11 #include<ctime>
12 #include<deque>
13 #include<iomanip>
14 #include<list>
15 #include<map>
16 #include<queue>
17 #include<set>
18 #include<stack>
19 #include<vector>
20 using namespace std;
21 typedef long long ll;
22 
23 const double PI=acos(-1.0);
24 const double eps=1e-6;
25 const ll mod=1e9+7;
26 const int inf=0x3f3f3f3f;
27 const int maxn=100+10;
28 #define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
29 
30 ll n,a[maxn],sum[maxn][maxn];
31 
32 void init()
33 {
34     int cnt=0;
35     for(int i=0;i<4*n;++i){
36         for(int j=0;j<=i;++j){
37             sum[j][i-j]=a[cnt];
38             cnt=(cnt+1)%n;
39         }
40     }
41     for(int i=0;i<2*n;i++){
42         for(int j=0;j<2*n;j++){
43             if(i> 0&&j> 0) sum[i][j]+=sum[i-1][j]+sum[i][j-1]-sum[i-1][j-1];
44             if(i==0&&j> 0) sum[i][j]+=sum[i][j-1];
45             if(i> 0&&j==0) sum[i][j]+=sum[i-1][j];
46         }
47     }
48 }
49 
50 ll GetAns(int x,int y)
51 {
52     ll ans=0;
53     ans+=(x/n)*(y/n)*sum[n-1][n-1];
54     ans+=(y/n)*sum[x%n][n-1];
55     ans+=(x/n)*sum[n-1][y%n];
56     ans+=sum[x%n][y%n];
57     return ans;
58 }
59 
60 int main()
61 {
62     int t;scanf("%d",&t);
63     while(t--){
64         scanf("%lld",&n);
65         for(int i=0;i<n;i++)
66             scanf("%lld",&a[i]);
67         init();
68         n=n*2;
69         int m;scanf("%d",&m);
70         for(int i=0;i<m;i++){
71             int x1,y1,x2,y2;
72             scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
73             ll ans=GetAns(x2,y2)-GetAns(x1-1,y2)-GetAns(x2,y1-1)+GetAns(x1-1,y1-1);//容斥
74             printf("%lld\n",ans);
75         }
76     }
77 }