[Bzoj4403] sequence statistics

Inserted a plate to make this figure represents +1, the problem is converted to the length of the sequence 1 ~ n, rl insertion plates (inclusive card may be, not limited to the number of each point plate), there are many kinds of programs, i.e., seeking $ \ sum_ {i = 1} ^ {n} c (rl + i, rl) $
considering this formula, supplemented on a c (rl + 1, rl + 1), the original formula = $ \ sum_ {I = 1} ^ {n} c (rl + i, rl) + c (rl + 1, rl + 1) -1 = \ sum_ {I = 2} ^ {n} c (rl + i, rl) + c (rl + 2, rl + 1) -1 = ...... = c (rl + n + 1, rl + 1) -1 $ , can be calculated by Lucas

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define mod 1000003
 4 int t,n,l,r,fac[mod],inv[mod];
 5 int c(int n,int m){
 6     if (n<m)return 0;
 7     return 1LL*fac[n]*inv[m]%mod*inv[n-m]%mod;
 8 }
 9 int main(){
10     scanf("%d",&t);
11     fac[0]=inv[0]=inv[1]=1;
12     for(int i=1;i<mod;i++)fac[i]=1LL*fac[i-1]*i%mod;
13     for(int i=2;i<mod;i++)inv[i]=1LL*(mod-mod/i)*inv[mod%i]%mod;
14     for(int i=2;i<mod;i++)inv[i]=1LL*inv[i-1]*inv[i]%mod;
15     while (t--){
16         scanf("%d%d%d",&n,&l,&r);
17         l=r-l+1;
18         printf("%lld\n",(1LL*c((n+l)/mod,l/mod)*c((n+l)%mod,l%mod)+mod-1)%mod);
19     }
20 } 

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