等比数列求和-分治法

等比数列求和-分治法

题目

$(1+p+p^2+...+p^c)modB$

因为等比数列直接用求和公式会出现分数形式，不能对分子和分母进行mod运算，再做除法
mod只对加、减、乘具有分配律

若 c为奇数

$$1+p+p^2+...+p^c$$

$$=（1+p+p^2+..+p^{\frac{c-1}{2}）+(p^{\frac{c+1}{2}}+p^{\frac{c+3}{2}}+...+p^c})$$

$$=(1+p+p^2+..+p^{\frac{c-1}{2}}）+p^{\frac{c+1}{2}}\ast (1+p+p^2+..+p^{\frac{c-1}{2}})$$

$$=(1+p^{\frac{c+1}{2}})\ast (1+p+p^2+..+p^{\frac{c-1}{2}})$$

$$=(1+p^{\frac{c+1}{2}})\ast sum(p,\frac{c-1}{2})$$

若 c为偶数

$$1+p+p^2+...+p^c$$

$$=（1+p+p^2+..+p^{\frac{c}{2}-1）+(p^{\frac{c}{2}}+p^{\frac{c}{2}+1}+...+p^c})$$

$$=(1+p+p^2+..+p^{\frac{c}{2}-1}）+p^{\frac{c}{2}}\ast (1+p+p^2+..+p^{\frac{c-1}{2}})+p^c$$

$$=(1+p^{\frac{c}{2}})\ast (1+p+p^2+..+p^{\frac{c-1}{2}})+p^c$$

$$=(1+p^{\frac{c}{2}})\ast sum(p,\frac{c-1}{2})+p^c$$

代码

ll ksm(ll a, ll b) {
    ll ans = 1;
    while(b) {
        if(b & 1) {
            ans *= a;
        }
        a *= a;
        b >>= 1;
    }
    return ans;
}

ll sum(ll a, ll b) {
    if(b == 0) return 1;
    if(b == 1) return a;
    if(b & 1) {
        return ((1 + ksm(a, (b + 1) / 2)) * sum(a, (b - 1) / 2));
    } else {
        return ((1 + ksm(a, b / 2) * sum(a, (b - 1) / 2)) + ksm(a, b));
    }
}