sun_qy @ 2024-01-03 16:18:34
#include <bits/stdc++.h>
using namespace std;
struct node {
int a[22500];
node() {
memset(a, 0, sizeof(a));
}
node(unsigned long long x) {
int i = 1;
while(x) {
a[i ++] = x % 10;
x /= 10;
}
a[0] = -- i;
}
void input() {
string s;
cin >> s;
a[0] = s.size();
for(int i = 1; i <= a[0]; i ++)
a[i] = s[a[0] - i] - '0';
}
bool zero() {
return a[0] == 1 && a[1] == 0;
}
void print() {
if(zero()) {
printf("0\n");
return;
}
int tmp = 0;
for(int i = a[0]; i >= 1; i --) {
if(a[i] || tmp) {
printf("%d", a[i]);
tmp = 1;
}
}
printf("\n");
}
};
inline node operator + (node x, node y) {
node res;
res.a[0] = max(x.a[0], y.a[0]);
for(int i = 1; i <= res.a[0]; i ++) {
res.a[i] += x.a[i] + y.a[i];
if(res.a[i] >= 10) res.a[i] -= 10, res.a[i + 1] ++;
}
if(res.a[res.a[0] + 1]) res.a[0] ++;
return res;
}
inline node operator * (node x, node y) {
int lx = x.a[0], ly = y.a[0];
node res;
if(x.zero()) return x;
if(y.zero()) return y;
res.a[0] = lx + ly;
for(int i = 1; i <= lx; i ++) {
for(int j = 1; j <= ly; j ++) {
res.a[i + j - 1] += ((x.a[i] * y.a[j]) % 10);
res.a[i + j] += ((x.a[i] * y.a[j]) / 10);
res.a[i + j] += res.a[i + j - 1] / 10;
res.a[i + j - 1] %= 10;
}
}
if(res.a[lx + ly]) res.a[0] ++;
return res;
}
inline node operator - (node x, node y) {
node res;
if(y.zero()) return x;
res.a[0] = max(x.a[0], y.a[0]);
for(int i = 1; i <= res.a[0]; i ++) {
res.a[i] += x.a[i] - y.a[i];
if(res.a[i] < 0) x.a[i + 1] --, res.a[i] += 10;
}
if(res.a[res.a[0]] == 0) res.a[0] --;
if(res.a[0] == 0) res.a[0] = 1, res.a[1] = 0;
return res;
}
node fac(node x) {
node res = 1;
for(; !x.zero(); x = x - 1) {
res = res * x;
}
return res;
}
int main() {
node t, res;
t.input();
while(!t.zero()) {
res = res + fac(t);
t = t - 1;
}
res.print();
return 0;
}注：其它ojAC了，数据范围比这个还大，对拍没有任何问题
by yuqixin2008 @ 2024-01-03 19:24:59
其实这个不用重载运算符就能过的其坤
by sun_qy @ 2024-01-04 20:05:49
@myxsb 其实我知道这个，但是我喜欢这个写法
by swuster27 @ 2024-01-30 22:38:37
高精度加法+高精度乘法
by swuster27 @ 2024-01-30 22:50:47
#include<iostream>
using namespace std;
static const int N = 100;
int a[N], b[N];
int main()
{
int n, len;
cin >> n;
a[0] = 1; b[0] = 1;
for (int i = 2; i <= n; i++)
{
//高精度乘法
for (int j = 0; j <N ; j++)
{
a[j] *= i;
}
for (int j = 0; j <N; j++)
{
a[j + 1] += a[j] / 10;
a[j] %= 10;
}
//高精度加法
for (len = 0; len <N; len++)
{
b[len] += a[len];
b[len + 1] += b[len] / 10;
b[len] %= 10;
}
}
//删除多余前导0
for (len=N; b[len] == 0; len--);
for (int i = len; i >= 0; i--)
{
cout << b[i];
}
cout << endl;
return 0;
}