UVAlive 6426 Count【读入】

短命女 2022-08-04 13:39 142阅读 0赞

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

--------------------

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;
    }