It’s a Mod, Mod, Mod, Mod World
时间限制: 2 Sec 内存限制: 128 MB题目描述
You are given multiple problems with three integers p, q, and n. Find 
. That is, the first n multiples of p, modulo q, summed. Note that the overall sum has no modulus.
. That is, the first n multiples of p, modulo q, summed. Note that the overall sum has no modulus.输入
Each input will begin with a line with a single integer W (1≤W≤105 ), which is the number of cases you must solve.
Each of the next W lines will contain three space-separated integers p, q and n (1≤p, q, n≤106 ),which are the parameters of the problem as described above.
Each of the next W lines will contain three space-separated integers p, q and n (1≤p, q, n≤106 ),which are the parameters of the problem as described above.
输出
Output W lines, each with the answer for a given instance of the problem, in the order that they appear in the input.
样例输入
3
2 7 2
1 4 5
3 8 10
样例输出
6
7
37
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 const int maxn=1e5+7; 5 const ll mod=1e9+7; 6 ll p,q,n; 7 //a 公差 b 首项 c 除数 n 项数 8 ll work(ll a,ll b,ll c,ll n){ 9 if (n<=0) return 0; 10 if (n==1) return b/c ; 11 ll t = 0; 12 t += a/c*(n-1)*n/2+ b/c*n; 13 a = a%c; 14 b = b%c; 15 if (a==0) return t; 16 return t+work(c,(a*n+b)%c,a,(a*n+b)/c); 17 } 18 int main() 19 { 20 int t; 21 scanf("%d",&t); 22 while(t--) 23 { 24 scanf("%lld%lld%lld",&p,&q,&n); 25 ll ans=(p*(n+1)*n/2-q*work(p,p,q,n)); 26 printf("%lld ",ans); 27 } 28 return 0; 29 }