明明这次又要出去旅游了,和上次不同的是,他这次要去宇宙探险!我们暂且不讨论他有多么NC,他又幻想了他应
该带一些什么东西。理所当然的,你当然要帮他计算携带N件物品的方案数。他这次又准备带一些受欢迎的食物,
如:蜜桃多啦,鸡块啦,承德汉堡等等当然,他又有一些稀奇古怪的限制:每种食物的限制如下:
承德汉堡:偶数个
可乐:0个或1个
鸡腿:0个,1个或2个
蜜桃多:奇数个
鸡块:4的倍数个
包子:0个,1个,2个或3个
土豆片炒肉:不超过一个。
面包:3的倍数个
注意,这里我们懒得考虑明明对于带的食物该怎么搭配着吃,也认为每种食物都是以‘个’为单位(反正是幻想嘛
),只要总数加起来是N就算一种方案。因此,对于给出的N,你需要计算出方案数,并对10007取模。
Solution
套路题,写出生成函数,并化简为
[frac{x}{(1-x)^4}
]
展开为幂级数,运用插板法计算答案,得
[C_{n+2}^3
]
(多此一举地写了高精)
#include <bits/stdc++.h>
using namespace std;
const int maxlen = 2005;
class HP {
public:
int len, s[maxlen];
HP() { (*this) = 0; }
HP(int inte) { (*this) = inte; }
HP(const char *str) { (*this) = str; }
friend ostream &operator<<(ostream &cout, const HP &x);
HP operator=(int inte);
HP operator=(const char *str);
HP operator=(const HP &b);
HP operator*(const HP &b);
HP operator+(const HP &b);
HP operator-(const HP &b);
HP operator/(const HP &b);
HP operator%(const HP &b);
bool operator<(const HP &b);
bool operator>(const HP &b);
int Compare(const HP &b);
};
ostream &operator<<(ostream &cout, const HP &x) {
for (int i = x.len; i >= 1; i--) cout << x.s[i];
return cout;
}
HP HP::operator=(const char *str) {
len = (int)strlen(str);
for (int i = 1; i <= len; i++) s[i] = str[len - i] - '0';
return *this;
}
HP HP::operator=(int inte) {
if (inte == 0) {
len = 1;
s[1] = 0;
return (*this);
}
for (len = 0; inte > 0;) {
s[++len] = inte % 10;
inte /= 10;
}
return *this;
}
HP HP::operator=(const HP &b) {
len = b.len;
for (int i = 1; i <= len; i++) s[i] = b.s[i];
return *this;
}
HP HP::operator*(const HP &b) {
int i, j;
HP c;
c.len = len + b.len;
for (i = 1; i <= c.len; i++) c.s[i] = 0;
for (i = 1; i <= len; i++)
for (j = 1; j <= b.len; j++) c.s[i + j - 1] += s[i] * b.s[j];
for (i = 1; i < c.len; i++) {
c.s[i + 1] += c.s[i] / 10;
c.s[i] %= 10;
}
while (c.s[i]) {
c.s[i + 1] = c.s[i] / 10;
c.s[i] %= 10;
i++;
}
while (i > 1 && !c.s[i]) i--;
c.len = i;
return c;
}
HP HP::operator+(const HP &b) {
int i;
HP c;
c.s[1] = 0;
for (i = 1; i <= len || i <= b.len || c.s[i]; i++) {
if (i <= len)
c.s[i] += s[i];
if (i <= b.len)
c.s[i] += b.s[i];
c.s[i + 1] = c.s[i] / 10;
c.s[i] %= 10;
}
c.len = i - 1;
if (c.len == 0)
c.len = 1;
return c;
}
HP HP::operator-(const HP &b) {
int i, j;
HP c;
for (i = 1, j = 0; i <= len; i++) {
c.s[i] = s[i] - j;
if (i <= b.len)
c.s[i] -= b.s[i];
if (c.s[i] < 0) {
j = 1;
c.s[i] += 10;
} else
j = 0;
}
c.len = len;
while (c.len > 1 && !c.s[c.len]) c.len--;
return c;
}
int HP::Compare(const HP &y) {
if (len > y.len)
return 1;
if (len < y.len)
return -1;
int i = len;
while ((i > 1) && (s[i] == y.s[i])) i--;
return s[i] - y.s[i];
}
bool HP::operator<(const HP &b) {
if (len < b.len)
return 1;
if (len > b.len)
return 0;
int i = len;
while ((i > 1) && (s[i] == b.s[i])) i--;
return s[i] < b.s[i];
}
bool HP::operator>(const HP &b) {
if (len > b.len)
return 1;
if (len < b.len)
return 0;
int i = len;
while ((i > 1) && (s[i] == b.s[i])) i--;
return s[i] > b.s[i];
}
HP HP::operator/(const HP &b) {
int i, j;
HP d(0), c;
for (i = len; i > 0; i--) {
if (!(d.len == 1 && d.s[1] == 0)) {
for (j = d.len; j > 0; j--) d.s[j + 1] = d.s[j];
++d.len;
}
d.s[1] = s[i];
c.s[i] = 0;
while ((j = d.Compare(b)) >= 0) {
d = d - b;
c.s[i]++;
if (j == 0)
break;
}
}
c.len = len;
while ((c.len > 1) && (c.s[c.len] == 0)) c.len--;
return c;
}
HP HP::operator%(const HP &b) {
int i, j;
HP d(0);
for (i = len; i > 0; i--) {
if (!(d.len == 1 && d.s[1] == 0)) {
for (j = d.len; j > 0; j--) d.s[j + 1] = d.s[j];
++d.len;
}
d.s[1] = s[i];
while ((j = d.Compare(b)) >= 0) {
d = d - b;
if (j == 0)
break;
}
}
return d;
}
char t[505];
signed main() {
HP n;
cin>>t;
n=t;
cout<<(n*(n+1)*(n+2))/6%10007;
}