2017 Multi-University Training Contest

公式：

上面的a|b表示a是b的约数，ai是每个质因数p分解后的项数

枚举每个x(l<=x<=r)

对于每个数暴力分解小于sqrt(r)的质数就好了， 因为大于sqrt(r)的质数最多只有一个，且对应的ai也只能为1

也就是说如果分解完那个数不是1的话就将d(x^k)乘上(k+1)

最后全加在一起

#include<stdio.h>#include<math.h>#define mod 998244353#define LL long longLL cnt, flag[1000005] = {1,1}, pri[1000005], val[1000005], d[1000005];int main(void){LL T, l, r, k, i, j, ans, sum;for(i=2;i<=1000001;i++){if(flag[i])continue;for(j=i*i;j<=1000001;j+=i)flag[j] = 1;pri[++cnt] = i;}scanf("%lld", &T);while(T--){ans = 0;scanf("%lld%lld%lld", &l, &r, &k);for(i=l;i<=r;i++)val[i-l] = i, d[i-l] = 1;for(i=1;pri[i]*pri[i]<=r;i++){sum = l%pri[i];j = (pri[i]-sum)%pri[i];for(;j<=r-l;j+=pri[i]){sum = 0;while(val[j]%pri[i]==0)sum++, val[j] /= pri[i];d[j] = d[j]*(sum*k+1)%mod;}}for(i=0;i<=r-l;i++){if(val[i]>1)d[i] = (d[i]*(k+1))%mod;ans = (ans+d[i])%mod;}printf("%lld\n", ans);}return 0;}