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;
}