project euler 48 Self powers 解决乘法爆long long

题目链接

求 $ 1¹⁺²2+\cdots + 1000^{1000} $ %1e10 的结果。

唯一的坑点是会爆longlong，所以用特殊的乘法。

#include <iostream>

#include <vector>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <complex>

#include <cmath>

#include <map>

#include <set>

#include <string>

#include <queue>

#include <stack>

#include <bitset>

using namespace std;

#define pb(x) push_back(x)

#define ll long long

#define mk(x, y) make_pair(x, y)

#define lson l, m, rt<<1

#define mem(a) memset(a, 0, sizeof(a))

#define rson m+1, r, rt<<1|1

#define mem1(a) memset(a, -1, sizeof(a))

#define mem2(a) memset(a, 0x3f, sizeof(a))

#define rep(i, n, a) for(int i = a; i<n; i++)

#define fi first

#define se second

typedef complex <double> cmx;

typedef pair<int, int> pll;

const double PI = acos(-1.0);

const double eps = 1e-8;

const ll mod = 1e10;

const int inf = 1061109567;

const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };

ll mul(ll a, ll b) {

    ll ret = 0;

    while(b) {

        if(b&1) {

            ret += a;

            if(ret >= mod)

                ret -= mod;

        }

        a += a;

        if(a >= mod)

            a -= mod;

        b >>= 1;

    }

    return ret;

}

ll pow(ll a, ll b) {

    ll ret = 1;

    while(b) {

        if(b&1) {

            ret = mul(ret, a)%mod;

        }

        a = mul(a, a)%mod;

        b >>= 1;

    }

    return ret;

}

int main()

{

    ll sum = 0;

    for(ll i = 1; i <= 1000; i++) {

        sum = (sum + pow(i, i))%mod;

    }

    cout<<sum<<endl;

    return 0;

}

巴特西

project euler 48 Self powers 解决乘法爆long long

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