Sol
设 (n=lfloorfrac{c}{a}
floor)
问题转化为求
[sum_{i=0}^{n}lfloorfrac{c-ax}{b}
floor+1=sum_{i=0}^{n}lfloorfrac{-ax+b+c}{b}
floor
]
考虑一般性的问题
设
[f(a,b,c,n)=sum_{i=0}^{n}lfloorfrac{ax+b}{c}
floor，c
e 0
]
- 若 (cle 0)，那么 (f(a,b,c,n)=f(-a,-b,-c,n))
- 若 (a<0) 或 (b<0)，那么
[f(a,b,c,n)=f(a~mod~c + c, b~mod~c + c, c, n) + frac{n(n + 1)}{2} (lfloorfrac{a}{c}
floor - 1) + (n + 1)(lfloorfrac{b}{c}
floor - 1)
]
- 若 (a>=c) 或 (b>=c)，那么
[f(a,b,c,n)=f(a~mod~c, b~mod~c, c, n) + frac{n(n + 1)}{2}lfloorfrac{a}{c}
floor + (n + 1)lfloorfrac{b}{c}
floor
]
- 最后 (0le a<c) 且 (0le b<c)
设 (m=lfloorfrac{an+b}{c} floor)
那么
[sum_{i=0}^{n}lfloorfrac{ai+b}{c}
floor=sum_{i=0}^{n}sum_{j=1}^{m}[lfloorfrac{ai+b}{c}
floorge j]=sum_{i=0}^{n}sum_{j=0}^{m-1}[lfloorfrac{ai+b}{c}
floorge j+1]
]
[=sum_{i=0}^{n}sum_{j=0}^{m-1}[aige cj+c-b]=sum_{i=0}^{n}sum_{j=0}^{m-1}[ai> cj+c-b-1]
]
[=sum_{i=0}^{n}sum_{j=0}^{m-1}[i> lfloorfrac{cj+c-b-1}{a}
floor]=sum_{i=0}^{m-1}(n-lfloorfrac{ci+c-b-1}{a}
floor)
]
[=nm-sum_{i=0}^{m-1}lfloorfrac{ci+c-b-1}{a}
floor=nm-f(c,c-b-a,a,m-1)
]
边界是 (a=0) 或者 (nle 1)
这个题直接代入就好了
# include <bits/stdc++.h>
using namespace std;
typedef long long ll;
inline ll Gcd(ll x, ll y) {
if (!x || !y) return x + y;
return !y ? x : Gcd(y, x % y);
}
inline ll Solve(ll a, ll b, ll c, ll n) {
if (!a) return (n + 1) * (b / c);
if (!n) return b / c;
if (n == 1) return (a + b) / c + b / c;
if (c < 0) return Solve(-a, -b, -c, n);
register ll d = abs(Gcd(Gcd(a, b), c));
a /= d, b /= d, c /= d;
if (a >= c || b >= c) return Solve(a % c, b % c, c, n) + n * (n + 1) / 2 * (a / c) + (n + 1) * (b / c);
if (a < 0 || b < 0) return Solve(a % c + c, b % c + c, c, n) + n * (n + 1) / 2 * (a / c - 1) + (n + 1) * (b / c - 1);
register ll m = (a * n + b) / c;
return n * m - Solve(c, c - b - 1, a, m - 1);
}
ll a, b, c, n;
int main() {
scanf("%lld%lld%lld", &a, &b, &c), n = c / a;
printf("%lld
", Solve(-a, c + b, b, n));
return 0;
}