【洛谷-P1009】解题报告(高精度)

原始题目

P1009 阶乘之和

题目大意

阶乘之和

解题思路

爆ll,需要高精度,上ACM大数模板或者

人生苦短,我用PY

解题代码

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#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;

#define rep(i, a, n) for (int i = a; i < n; ++i)
#define per(i, a, n) for (int i = n - 1; i >= a; --i)
#define pb push_back

#define MAXN 9999
#define MAXSIZE 10
#define DLEN 4

class BigNum {
private:
int a[500]; //可以控制大数的位数
int len; //大数长度
public:
BigNum()
{
len = 1;
memset(a, 0, sizeof(a));
} //构造函数
BigNum(const int); //将一个int类型的变量转化为大数
BigNum(const char*); //将一个字符串类型的变量转化为大数
BigNum(const BigNum&); //拷贝构造函数
BigNum& operator=(const BigNum&); //重载赋值运算符,大数之间进行赋值运算

friend istream& operator>>(istream&, BigNum&); //重载输入运算符
friend ostream& operator<<(ostream&, BigNum&); //重载输出运算符

BigNum operator+(const BigNum&) const; //重载加法运算符,两个大数之间的相加运算
BigNum operator-(const BigNum&) const; //重载减法运算符,两个大数之间的相减运算
BigNum operator*(const BigNum&)const; //重载乘法运算符,两个大数之间的相乘运算
BigNum operator/(const int&) const; //重载除法运算符,大数对一个整数进行相除运算

BigNum operator^(const int&) const; //大数的n次方运算
int operator%(const int&) const; //大数对一个int类型的变量进行取模运算
bool operator>(const BigNum& T) const; //大数和另一个大数的大小比较
bool operator>(const int& t) const; //大数和一个int类型的变量的大小比较

void print(); //输出大数
};
BigNum::BigNum(const int b) //将一个int类型的变量转化为大数
{
int c, d = b;
len = 0;
memset(a, 0, sizeof(a));
while (d > MAXN) {
c = d - (d / (MAXN + 1)) * (MAXN + 1);
d = d / (MAXN + 1);
a[len++] = c;
}
a[len++] = d;
}
BigNum::BigNum(const char* s) //将一个字符串类型的变量转化为大数
{
int t, k, index, l, i;
memset(a, 0, sizeof(a));
l = strlen(s);
len = l / DLEN;
if (l % DLEN)
len++;
index = 0;
for (i = l - 1; i >= 0; i -= DLEN) {
t = 0;
k = i - DLEN + 1;
if (k < 0)
k = 0;
for (int j = k; j <= i; j++)
t = t * 10 + s[j] - '0';
a[index++] = t;
}
}
BigNum::BigNum(const BigNum& T)
: len(T.len) //拷贝构造函数
{
int i;
memset(a, 0, sizeof(a));
for (i = 0; i < len; i++)
a[i] = T.a[i];
}
BigNum& BigNum::operator=(const BigNum& n) //重载赋值运算符,大数之间进行赋值运算
{
int i;
len = n.len;
memset(a, 0, sizeof(a));
for (i = 0; i < len; i++)
a[i] = n.a[i];
return *this;
}
istream& operator>>(istream& in, BigNum& b) //重载输入运算符
{
char ch[MAXSIZE * 4];
int i = -1;
in >> ch;
int l = strlen(ch);
int count = 0, sum = 0;
for (i = l - 1; i >= 0;) {
sum = 0;
int t = 1;
for (int j = 0; j < 4 && i >= 0; j++, i--, t *= 10) {
sum += (ch[i] - '0') * t;
}
b.a[count] = sum;
count++;
}
b.len = count++;
return in;
}
ostream& operator<<(ostream& out, BigNum& b) //重载输出运算符
{
int i;
cout << b.a[b.len - 1];
for (i = b.len - 2; i >= 0; i--) {
cout.width(DLEN);
cout.fill('0');
cout << b.a[i];
}
return out;
}

BigNum BigNum::operator+(const BigNum& T) const //两个大数之间的相加运算
{
BigNum t(*this);
int i, big; //位数
big = T.len > len ? T.len : len;
for (i = 0; i < big; i++) {
t.a[i] += T.a[i];
if (t.a[i] > MAXN) {
t.a[i + 1]++;
t.a[i] -= MAXN + 1;
}
}
if (t.a[big] != 0)
t.len = big + 1;
else
t.len = big;
return t;
}
BigNum BigNum::operator-(const BigNum& T) const //两个大数之间的相减运算
{
int i, j, big;
bool flag;
BigNum t1, t2;
if (*this > T) {
t1 = *this;
t2 = T;
flag = 0;
} else {
t1 = T;
t2 = *this;
flag = 1;
}
big = t1.len;
for (i = 0; i < big; i++) {
if (t1.a[i] < t2.a[i]) {
j = i + 1;
while (t1.a[j] == 0)
j++;
t1.a[j--]--;
while (j > i)
t1.a[j--] += MAXN;
t1.a[i] += MAXN + 1 - t2.a[i];
} else
t1.a[i] -= t2.a[i];
}
t1.len = big;
while (t1.a[t1.len - 1] == 0 && t1.len > 1) {
t1.len--;
big--;
}
if (flag)
t1.a[big - 1] = 0 - t1.a[big - 1];
return t1;
}

BigNum BigNum::operator*(const BigNum& T) const //两个大数之间的相乘运算
{
BigNum ret;
int i, j, up;
int temp, temp1;
for (i = 0; i < len; i++) {
up = 0;
for (j = 0; j < T.len; j++) {
temp = a[i] * T.a[j] + ret.a[i + j] + up;
if (temp > MAXN) {
temp1 = temp - temp / (MAXN + 1) * (MAXN + 1);
up = temp / (MAXN + 1);
ret.a[i + j] = temp1;
} else {
up = 0;
ret.a[i + j] = temp;
}
}
if (up != 0)
ret.a[i + j] = up;
}
ret.len = i + j;
while (ret.a[ret.len - 1] == 0 && ret.len > 1)
ret.len--;
return ret;
}
BigNum BigNum::operator/(const int& b) const //大数对一个整数进行相除运算
{
BigNum ret;
int i, down = 0;
for (i = len - 1; i >= 0; i--) {
ret.a[i] = (a[i] + down * (MAXN + 1)) / b;
down = a[i] + down * (MAXN + 1) - ret.a[i] * b;
}
ret.len = len;
while (ret.a[ret.len - 1] == 0 && ret.len > 1)
ret.len--;
return ret;
}
int BigNum::operator%(const int& b) const //大数对一个int类型的变量进行取模运算
{
int i, d = 0;
for (i = len - 1; i >= 0; i--) {
d = ((d * (MAXN + 1)) % b + a[i]) % b;
}
return d;
}
BigNum BigNum::operator^(const int& n) const //大数的n次方运算
{
BigNum t, ret(1);
int i;
if (n < 0)
exit(-1);
if (n == 0)
return 1;
if (n == 1)
return *this;
int m = n;
while (m > 1) {
t = *this;
for (i = 1; i << 1 <= m; i <<= 1) {
t = t * t;
}
m -= i;
ret = ret * t;
if (m == 1)
ret = ret * (*this);
}
return ret;
}
bool BigNum::operator>(const BigNum& T) const //大数和另一个大数的大小比较
{
int ln;
if (len > T.len)
return true;
else if (len == T.len) {
ln = len - 1;
while (a[ln] == T.a[ln] && ln >= 0)
ln--;
if (ln >= 0 && a[ln] > T.a[ln])
return true;
else
return false;
} else
return false;
}
bool BigNum::operator>(const int& t) const //大数和一个int类型的变量的大小比较
{
BigNum b(t);
return *this > b;
}

void BigNum::print() //输出大数
{
int i;
cout << a[len - 1];
for (i = len - 2; i >= 0; i--) {
cout.width(DLEN);
cout.fill('0');
cout << a[i];
}
cout << endl;
}

int n;
int main()
{
ios::sync_with_stdio(false);
while (cin >> n) {
BigNum ans = 0;
for (int i = 1; i <= n; ++i) {
BigNum temp = 1;
rep(j, 1, i + 1)
{
temp = temp * j;
}
ans = ans + temp;
}
ans.print();
}
}
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num = int(input())
sum = 0
for i in range(1,num+1):
cnt = 1
for j in range(1,i+1):
cnt*=j
sum+=cnt
print(sum)

收获与总结