[洛谷P3935]Calculating

题目大意：设把$x$分解质因数的结果为$x=p_1^{k_1}p_2^{k_2}cdots p_n^{k_n}$，令$f(x)=(k_1+1)(k_2+1)cdots (k_n+1)$，求$sumlimits_{i=l}^r f(i)(1leqslant lleqslant 10^{14},1leqslant rleqslant 1.6 imes10^{14},r-l>10^{14}$

题解：可知$f(x)$为$x$的因数个数，可以把$sumlimits_{i=l}^rf(i)$拆成$sumlimits_{i=1}^rf(i)-sumlimits_{i=1}^lf(i)$。
$$
defdsum{displaystylesumlimits}
defdprod{displaystyleprodlimits}
egin{align*}
f(p)&=dprod_{i=1}^n(k_{p,i}+1)\
    &=dsum_{i=1}^n[i|p]\
end{align*}\
带回原式
$$

$$
defdsum{displaystylesumlimits}
defdprod{displaystyleprodlimits}
egin{align*}
令g(p)&=dsum_{x=1}^pf(x)\
    &=dsum_{x=1}^pdsum_{i=1}^x[i|x]\
    &=dsum_{i=1}^piglfloordfrac p iig floor\
end{align*}
$$

整除分块即可。

卡点：1.读入时忘记开$long;long$



C++ Code：

#include <cstdio>
using namespace std;
const int mod = 998244353;
long long l, r;
long long solve(long long n) {
    long long ans = 0, l, r;
    for (l = 1; l <= n; l = r + 1) {
        r = n / (n / l);
        ans = (ans + (r - (l - 1)) * (n / l)) % mod;
    }
    return ans;
}
int main() {
    scanf("%lld%lld", &l, &r);
    printf("%lld
", (solve(r) - solve(l - 1) + mod) % mod);
    return 0;
}