标签：script beginning first main blog rip val printf tin
You are given an N × N matrix. At the beginning every element is 0. Write a program supporting 2 operations:
1. Add x y value: Add value to the element Axy. (Subscripts starts from 0
2. Sum x1 y1 x2 y1: Return the sum of every element Axy for x1 ≤ x ≤ x2, y1 ≤ y ≤ y2.
Input
The first line contains 2 integers N and M, the size of the matrix and the number of operations.
Each of the following M line contains an operation.
1 ≤ N ≤ 1000, 1 ≤ M ≤ 100000
For each Add operation: 0 ≤ x < N, 0 ≤ y < N, -1000000 ≤ value ≤ 1000000
For each Sum operation: 0 ≤ x1 ≤ x2 < N, 0 ≤ y1 ≤ y2 < N
Output
For each Sum operation output a non-negative number denoting the sum modulo 109+7.
Sample Input
5 8 Add 0 0 1 Sum 0 0 1 1 Add 1 1 1 Sum 0 0 1 1 Add 2 2 1 Add 3 3 1 Add 4 4 -1 Sum 0 0 4 4
Sample Output
1 2 3
二维的树状数组，做法与一维类似，注意求区间时加减范围。
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 6 using namespace std; 7 8 const int maxn=1005; 9 int g[maxn][maxn],n; 10 11 void update(int x,int y,int v) 12 { 13 int j=y; 14 for(;x<=n;x+=x&(-x)) 15 for(y=j;y<=n;y+=y&(-y)) 16 { 17 g[x][y]+=v; 18 g[x][y]%=1000000007; 19 } 20 } 21 22 long long getsum(int x,int y) 23 { 24 long long sum=0; 25 int j=y; 26 for(;x>0;x-=x&(-x)) 27 for(y=j;y>0;y-=y&(-y)) 28 { 29 sum+=g[x][y]; 30 sum%=1000000007; 31 } 32 return sum; 33 } 34 35 int main() 36 { 37 int m,x1,y1,x2,y2,v; 38 while(~scanf("%d%d",&n,&m)) 39 { 40 memset(g,0,sizeof(g)); 41 char s[5]; 42 for(int i=0;i<m;i++) 43 { 44 cin>>s; 45 if(s[0]==‘A‘) 46 { 47 scanf("%d%d%d",&x1,&y1,&v); 48 update(x1+1,y1+1,v); 49 } 50 else 51 { 52 scanf("%d%d%d%d",&x1,&y1,&x2,&y2); 53 long long ans=0; 54 ans+=getsum(x2+1,y2+1)%(1000000007); 55 ans-=getsum(x1,y2+1)%(1000000007); 56 ans-=getsum(x2+1,y1)%(1000000007); 57 ans+=getsum(x1,y1)%(1000000007); 58 ans%=1000000007; 59 if(ans<0) 60 ans+=1000000007; 61 printf("%lld\n",ans); 62 } 63 } 64 } 65 66 67 return 0; 68 }
标签：script beginning first main blog rip val printf tin
原文地址：http://www.cnblogs.com/xibeiw/p/7371393.html
