UVAlive 6426 Count【读入】

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You have:
• A matrix of natural numbers, with the property that all rows and all columns are sorted in ascending order (i.e. $A[i,j] ≥ A[i−1,j]$ and $A[i,j] ≥ A[i,j −1]$for all i, j)
• One or several pairs of numbers $(X,Y )$ with the property that $Y ≥ X$. For each $(X,Y )$ pair, count how many numbers from the matrix are greater than or equal to $X$ but smaller than or equal to $Y$ .

Input

The input file is a binary file containing 32-bit integer numbers. The input file consists of:
• One integer N representing the number of rows (no more than 10000)
• One integer M representing the number of columns (no more than 10000)
• $N ×M$ integers, representing the values from the matrix, row by row
• An unspecified number of integers, representing the$(X,Y )$pairs, one pair at a time. There will be at least one pair and at most 100 pairs in the file — and there will not be an incomplete pair at the end of the file.

Output

For each pair you should write to standard output a value representing how many numbers in the matrix are greater than or equal to $X$ but smaller than or equal to $Y$ .
Note: The sample input is here in text form, not binary, for obvious reasons
Sample Input

2 4
1 5 10 10
2 10 20 99
10 99
2 9
100 1000
10 10
Sample Output
5
2
0
3

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大水题
但是就是读入的时候有点迷，要求二进制读入。
语法问题吧？

#include<bits/stdc++.h>
#define MAXN 5010
#define MAXE 500000
typedef long long LL;
using namespace std;
int A[10005][10005];
bool read(int *a, int n){return fread(a,4,n,stdin);
}
int main(){int n, m;read(&n,1); read(&m,1);for(int i=0;i<n;i++){read(A[i],m);}int l,r;while(read(&l,1)&&read(&r,1)){int ans=0;for(int i=0;i<n;i++){int lpos=lower_bound(A[i],A[i]+m,l)-A[i];int rpos=upper_bound(A[i],A[i]+m,r)-A[i];if(rpos>lpos) ans+=rpos-lpos;}printf("%d\n",ans);}return 0;
}

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