How many hdu 1799 cycles?

How many cycles?
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission (s): 2096 Accepted Submission (s): 755

 

Problem Description
We know that in programming, we often need to consider the time complexity, especially for the loop. For example,
If
For (I = 1; I <= n; I ++) OP;
N OP operations are performed.
Fori = 1; I <= n; I ++)
For (j = I + 1; j <= n; j ++) OP;
Then, n * (n-1)/2 OP operations are performed.
Now we know that there are m-layer for loop operations, and the starting value of each for variable is the starting value of the previous variable + 1 (the starting value of the first variable is 1 ), the end value is an input n, and the total calculation amount of the OP is asked.

 

Input
There are T group cases, T <= 10000. Each case has two integers, m and n, 0 <m <= 2000, 0 <n <=.
 

Output
For each case, output a value, indicating the total calculation amount. Maybe this number is large, you only need to output the remainder except 1007.
 

Sample Input
2
1 3
2 3

Sample Output
3
3. This question is a bit of a dynamic planning taste. It is based on the former, and the poor status is transferred.
0 1 2 3 4
1 1 1 1
2 0 1 2 3
3 0 0 1 3
4 0 0 0 1

See the table does not, according to the table unfortunately concluded that the number of j in the I layer has dp [I] [j] = dp [I] [J-1] + dp [I-1] [J-1]
[Cpp] view plaincopyprint? # Include <iostream>
# Include <cstdio>
# Include <cstring>
Using namespace std;
# Define N 2001
# Define M 1007
Int f [N] [N];
Int main (){
Int I, j, n, m, T;
For (I = 1; I <N; I ++) f [1] [I] = 1;
For (I = 2; I <N; I ++) // layer I
For (j = I; j <N; j ++) {// Number of j
F [I] [j] = f [I] [J-1] + f [I-1] [J-1];
If (f [I] [j]> = M) f [I] [j] % = M;
}
Scanf ("% d", & T );
While (T --){
Scanf ("% d", & m, & n );
For (I = m, j = 0; I <= n; I ++ ){
J + = f [m] [I];
If (j> = M) j % = M;
}
Printf ("% d \ n", j );
}
 
Return 0;
}

# Include <iostream>
# Include <cstdio>
# Include <cstring>
Using namespace std;
# Define N 2001
# Define M 1007
Int f [N] [N];
Int main (){
Int I, j, n, m, T;
For (I = 1; I <N; I ++) f [1] [I] = 1;
For (I = 2; I <N; I ++) // layer I
For (j = I; j <N; j ++) {// Number of j
F [I] [j] = f [I] [J-1] + f [I-1] [J-1];
If (f [I] [j]> = M) f [I] [j] % = M;
}
Scanf ("% d", & T );
While (T --){
Scanf ("% d", & m, & n );
For (I = m, j = 0; I <= n; I ++ ){
J + = f [m] [I];
If (j> = M) j % = M;
}
Printf ("% d \ n", j );
}

Return 0;
}