以为是个大水题。。。
结果写了两小时
我数学真差.jpg
容易发现式子可以化成

然后拆开可以得到(四个)五个式子，分别计算即可…
（懒得写公式了，自己推一下吧
然后你就能发现死活过不去…
原因是啥呢…给的这个模数竟然不是个质数。
那么考虑用exgcd来求解逆元就行了
c++代码如下： #include #define rep(i,x,y) for(register int i = x ;i <= y;++ i) #define repd(i,x,y) for(register int i = x; i >= y ;-- i) using namespace std; typedef long long ll; templateinline void read(T&x) { x = 0;char c;int sign = 1; do { c = getchar(); if(c == '-') sign = -1; }while(!isdigit(c)); do { x = x * 10 + c - '0'; c = getchar(); }while(isdigit(c)); x *= sign; } const int mod = 19940417; ll n,m,n1,n2; ll ans,cal1,cal2,cal3; inline ll get(ll x) {return ((x+1)*x%mod)*n1%mod;} inline ll get2(ll x) {return ((2*x+1)*(x+1)%mod*x%mod)*n2%mod;} void exgcd(ll a,ll b,ll&x,ll&y) { if(!b) { x = 1;y = 0; return; } exgcd(b,a%b,y,x); y -= a/b*x; } int main() { read(n); read(m); ll x; exgcd(1ll*2,mod,n1,x);n1 = (n1+mod)%mod; exgcd(1ll*6,mod,n2,x);n2 = (n2+mod)%mod; ans = n * m % mod * n %mod * m % mod; for(int i = 1,lst;i <= n; i = lst + 1) { lst = n/(n/i); cal1 = (cal1 + (get(lst) - get(i-1)) * (n/i)%mod)% mod; } for(int i = 1,lst;i <= m; i = lst + 1) { lst = m/(m/i); cal2 = (cal2 + (get(lst) - get(i-1)) * (m/i)%mod)% mod; } cal3 = min(n,m) * n % mod * m %mod ; for(int i = 1,lst;i <= min(n,m); i = lst + 1) { lst = min(m/(m/i),n/(n/i)); cal3 = (cal3 + (mod - m*(get(lst) - get(i-1))%mod * (n/i) % mod))% mod; cal3 = (cal3 + (mod - n*(get(lst) - get(i-1))%mod * (m/i) %mod)) % mod; cal3 = (cal3 + (get2(lst) - get2(i - 1)+mod)%mod * (n/i)%mod * (m/i) %mod) % mod; } printf("%lld ",((ans - cal3 - (m*m%mod*cal1%mod) - (n*n%mod*cal2%mod) + cal1 * cal2%mod) %mod + mod )%mod); return 0; }