烷烃同分异构体数目的计算

烷烃同分异构体数目的计算

相信大家在学习高中的有机化学时，老师一定会给你讲过一类找烷烃同分异构体的题目吧？

而且老师还会说这种题只能靠动手画图，没有数学公式来计算，只有画的做到不重复不遗漏考试才能得到分。不少人为此绞尽脑汁。那么如果我们想知道几十个C原子能组成多少同分异构体怎么办呢？要一个一个地画吗？那估计穷尽一生也画不完。因为烷烃同分异构的数目变化和C原子的数目是不成线性变化的。

那么就没有计算的方法了吗？

其实不然，我么可以用Polya计数定理来推算烷烃同分异构的数目。
当然在这里我就不赘述数学上的证明了，有兴趣的小伙伴可以自行查阅有关书籍。

这个代码段通过高精度模版实现了任意碳原子数目构成同分异构体数目的计算。
它把同分异构体数目的计算巧妙等价转换成了烷基数目的计算，大大简化了计算难度。
然而，由于该算法存在不小的时间复杂度，也许1000以上个C原子组成的烷烃会计算很长时间。

源代码如下 注意：只能计算出链结构的烷烃

#include <iostream>
#include <cstring>#define DIGIT    4
#define DEPTH    10000
#define MAX     100
using namespace std;
typedef int bignum_t[MAX + 1];
int read(bignum_t a, istream& is = cin) {char buf[MAX * DIGIT + 1], ch;int i, j;memset((void*)a, 0, sizeof(bignum_t));if (!(is >> buf))    return 0;for (a[0] = strlen(buf), i = a[0] / 2 - 1; i >= 0; i--)ch = buf[i], buf[i] = buf[a[0] - 1 - i], buf[a[0] - 1 - i] = ch;for (a[0] = (a[0] + DIGIT - 1) / DIGIT, j = strlen(buf); j < a[0] * DIGIT; buf[j++] = '0');for (i = 1; i <= a[0]; i++)for (a[i] = 0, j = 0; j < DIGIT; j++)a[i] = a[i] * 10 + buf[i * DIGIT - 1 - j] - '0';for (; !a[a[0]] && a[0] > 1; a[0]--);return 1;
}
void write(const bignum_t a, ostream& os = cout) {int i, j;for (os << a[i = a[0]], i--; i; i--)for (j = DEPTH / 10; j; j /= 10)os << a[i] / j % 10;
}
int comp(const bignum_t a, const bignum_t b) {int i;if (a[0] != b[0])return a[0] - b[0];for (i = a[0]; i; i--)if (a[i] != b[i])return a[i] - b[i];return 0;
}
int comp(const bignum_t a, const int b) {int c[12] = { 1 };for (c[1] = b; c[c[0]] >= DEPTH; c[c[0] + 1] = c[c[0]] / DEPTH, c[c[0]] %= DEPTH, c[0]++);return comp(a, c);
}
int comp(const bignum_t a, const int c, const int d, const bignum_t b) {int i, t = 0, O = -DEPTH * 2;if (b[0] - a[0] < d && c)return 1;for (i = b[0]; i > d; i--) {t = t * DEPTH + a[i - d] * c - b[i];if (t > 0) return 1;if (t < O) return 0;}for (i = d; i; i--) {t = t * DEPTH - b[i];if (t > 0) return 1;if (t < O) return 0;}return t > 0;
}
void add(bignum_t a, const bignum_t b) {int i;for (i = 1; i <= b[0]; i++)if ((a[i] += b[i]) >= DEPTH)a[i] -= DEPTH, a[i + 1]++;if (b[0] >= a[0])a[0] = b[0];elsefor (; a[i] >= DEPTH && i < a[0]; a[i] -= DEPTH, i++, a[i]++);a[0] += (a[a[0] + 1] > 0);
}
void add(bignum_t a, const int b) {int i = 1;for (a[1] += b; a[i] >= DEPTH && i < a[0]; a[i + 1] += a[i] / DEPTH, a[i] %= DEPTH, i++);for (; a[a[0]] >= DEPTH; a[a[0] + 1] = a[a[0]] / DEPTH, a[a[0]] %= DEPTH, a[0]++);
}
void sub(bignum_t a, const bignum_t b) {int i;for (i = 1; i <= b[0]; i++)if ((a[i] -= b[i]) < 0)a[i + 1]--, a[i] += DEPTH;for (; a[i] < 0; a[i] += DEPTH, i++, a[i]--);for (; !a[a[0]] && a[0] > 1; a[0]--);
}
void sub(bignum_t a, const int b) {int i = 1;for (a[1] -= b; a[i] < 0; a[i + 1] += (a[i] - DEPTH + 1) / DEPTH, a[i] -= (a[i] - DEPTH + 1) / DEPTH * DEPTH, i++);for (; !a[a[0]] && a[0] > 1; a[0]--);
}
void sub(bignum_t a, const bignum_t b, const int c, const int d) {int i, O = b[0] + d;for (i = 1 + d; i <= O; i++)if ((a[i] -= b[i - d] * c) < 0)a[i + 1] += (a[i] - DEPTH + 1) / DEPTH, a[i] -= (a[i] - DEPTH + 1) / DEPTH * DEPTH;for (; a[i] < 0; a[i + 1] += (a[i] - DEPTH + 1) / DEPTH, a[i] -= (a[i] - DEPTH + 1) / DEPTH * DEPTH, i++);for (; !a[a[0]] && a[0] > 1; a[0]--);
}
void mul(bignum_t c, const bignum_t a, const bignum_t b) {int i, j;memset((void*)c, 0, sizeof(bignum_t));for (c[0] = a[0] + b[0] - 1, i = 1; i <= a[0]; i++)for (j = 1; j <= b[0]; j++)if ((c[i + j - 1] += a[i] * b[j]) >= DEPTH)c[i + j] += c[i + j - 1] / DEPTH, c[i + j - 1] %= DEPTH;for (c[0] += (c[c[0] + 1] > 0); !c[c[0]] && c[0] > 1; c[0]--);
}
void mul(bignum_t a, const int b) {int i;for (a[1] *= b, i = 2; i <= a[0]; i++) {a[i] *= b;if (a[i - 1] >= DEPTH)a[i] += a[i - 1] / DEPTH, a[i - 1] %= DEPTH;}for (; a[a[0]] >= DEPTH; a[a[0] + 1] = a[a[0]] / DEPTH, a[a[0]] %= DEPTH, a[0]++);for (; !a[a[0]] && a[0] > 1; a[0]--);
}
void mul(bignum_t b, const bignum_t a, const int c, const int d) {int i;memset((void*)b, 0, sizeof(bignum_t));for (b[0] = a[0] + d, i = d + 1; i <= b[0]; i++)if ((b[i] += a[i - d] * c) >= DEPTH)b[i + 1] += b[i] / DEPTH, b[i] %= DEPTH;for (; b[b[0] + 1]; b[0]++, b[b[0] + 1] = b[b[0]] / DEPTH, b[b[0]] %= DEPTH);for (; !b[b[0]] && b[0] > 1; b[0]--);
}
void div(bignum_t c, bignum_t a, const bignum_t b) {int h, l, m, i;memset((void*)c, 0, sizeof(bignum_t));c[0] = (b[0] < a[0] + 1) ? (a[0] - b[0] + 2) : 1;for (i = c[0]; i; sub(a, b, c[i] = m, i - 1), i--)for (h = DEPTH - 1, l = 0, m = (h + l + 1) >> 1; h > l; m = (h + l + 1) >> 1)if (comp(b, m, i - 1, a)) h = m - 1;else l = m;for (; !c[c[0]] && c[0] > 1; c[0]--);c[0] = c[0] > 1 ? c[0] : 1;
}
void div(bignum_t a, const int b, int& c) {int i;for (c = 0, i = a[0]; i; c = c * DEPTH + a[i], a[i] = c / b, c %= b, i--);for (; !a[a[0]] && a[0] > 1; a[0]--);
}
void sqrt(bignum_t b, bignum_t a) {int h, l, m, i;memset((void*)b, 0, sizeof(bignum_t));for (i = b[0] = (a[0] + 1) >> 1; i; sub(a, b, m, i - 1), b[i] += m, i--)for (h = DEPTH - 1, l = 0, b[i] = m = (h + l + 1) >> 1; h > l; b[i] = m = (h + l + 1) >> 1)if (comp(b, m, i - 1, a)) h = m - 1;else l = m;for (; !b[b[0]] && b[0] > 1; b[0]--);for (i = 1; i <= b[0]; b[i++] >>= 1);
}
int length(const bignum_t a) {int t, ret;for (ret = (a[0] - 1) * DIGIT, t = a[a[0]]; t; t /= 10, ret++);return ret > 0 ? ret : 1;
}
int digit(const bignum_t a, const int b) {int i, ret;for (ret = a[(b - 1) / DIGIT + 1], i = (b - 1) % DIGIT; i; ret /= 10, i--);return ret % 10;
}
int zeronum(const bignum_t a) {int ret, t;for (ret = 0; !a[ret + 1]; ret++);for (t = a[ret + 1], ret *= DIGIT; !(t % 10); t /= 10, ret++);return ret;
}
void comp(int* a, const int l, const int h, const int d) {int i, j, t;for (i = l; i <= h; i++)for (t = i, j = 2; t > 1; j++)while (!(t % j))a[j] += d, t /= j;
}
void convert(int* a, const int h, bignum_t b) {int i, j, t = 1;memset(b, 0, sizeof(bignum_t));for (b[0] = b[1] = 1, i = 2; i <= h; i++)if (a[i])for (j = a[i]; j; t *= i, j--)if (t * i > DEPTH)mul(b, t), t = 1;mul(b, t);
}
void combination(bignum_t a, int m, int n) {int* t = new int[m + 1];memset((void*)t, 0, sizeof(int) * (m + 1));comp(t, n + 1, m, 1);comp(t, 2, m - n, -1);convert(t, m, a);delete[]t;
}
void permutation(bignum_t a, int m, int n) {int i, t = 1;memset(a, 0, sizeof(bignum_t));a[0] = a[1] = 1;for (i = m - n + 1; i <= m; t *= i++)if (t * i > DEPTH)mul(a, t), t = 1;mul(a, t);
}#define SGN(x) ((x)>0?1:((x)<0?-1:0))
#define ABS(x) ((x)>0?(x):-(x))int read(bignum_t a, int& sgn, istream& is = cin) {char str[MAX * DIGIT + 2], ch, * buf;int i, j;memset((void*)a, 0, sizeof(bignum_t));if (!(is >> str)) return 0;buf = str, sgn = 1;if (*buf == '-') sgn = -1, buf++;for (a[0] = strlen(buf), i = a[0] / 2 - 1; i >= 0; i--)ch = buf[i], buf[i] = buf[a[0] - 1 - i], buf[a[0] - 1 - i] = ch;for (a[0] = (a[0] + DIGIT - 1) / DIGIT, j = strlen(buf); j < a[0] * DIGIT; buf[j++] = '0');for (i = 1; i <= a[0]; i++)for (a[i] = 0, j = 0; j < DIGIT; j++)a[i] = a[i] * 10 + buf[i * DIGIT - 1 - j] - '0';for (; !a[a[0]] && a[0] > 1; a[0]--);if (a[0] == 1 && !a[1]) sgn = 0;return 1;
}
struct bignum {bignum_t num;int sgn;
public:inline bignum() { memset(num, 0, sizeof(bignum_t)); num[0] = 1; sgn = 0; }inline int operator!() { return num[0] == 1 && !num[1]; }inline bignum& operator=(const bignum& a) { memcpy(num, a.num, sizeof(bignum_t)); sgn = a.sgn; return *this; }inline bignum& operator=(const int a) { memset(num, 0, sizeof(bignum_t)); num[0] = 1; sgn = SGN(a); add(num, sgn * a); return *this; };inline bignum& operator+=(const bignum& a) {if (sgn == a.sgn)add(num, a.num); else if (sgn && a.sgn) {int ret = comp(num, a.num); if (ret > 0)sub(num, a.num); else if (ret < 0) {bignum_t t;memcpy(t, num, sizeof(bignum_t)); memcpy(num, a.num, sizeof(bignum_t)); sub(num, t); sgn = a.sgn;}else memset(num, 0, sizeof(bignum_t)), num[0] = 1, sgn = 0;}else if (!sgn)memcpy(num, a.num, sizeof(bignum_t)), sgn = a.sgn; return *this;}inline bignum& operator+=(const int a) {if (sgn * a > 0)add(num, ABS(a)); else if (sgn && a) {int ret = comp(num, ABS(a)); if (ret > 0)sub(num, ABS(a)); else if (ret < 0) {bignum_t t;memcpy(t, num, sizeof(bignum_t)); memset(num, 0, sizeof(bignum_t)); num[0] = 1; add(num, ABS(a)); sgn = -sgn; sub(num, t);}else memset(num, 0, sizeof(bignum_t)), num[0] = 1, sgn = 0;}else if (!sgn)sgn = SGN(a), add(num, ABS(a)); return *this;}inline bignum operator+(const bignum& a) { bignum ret; memcpy(ret.num, num, sizeof(bignum_t)); ret.sgn = sgn; ret += a; return ret; }inline bignum operator+(const int a) { bignum ret; memcpy(ret.num, num, sizeof(bignum_t)); ret.sgn = sgn; ret += a; return ret; }inline bignum& operator-=(const bignum& a) {if (sgn * a.sgn < 0)add(num, a.num); else if (sgn && a.sgn) {int ret = comp(num, a.num); if (ret > 0)sub(num, a.num); else if (ret < 0) {bignum_t t;memcpy(t, num, sizeof(bignum_t)); memcpy(num, a.num, sizeof(bignum_t)); sub(num, t); sgn = -sgn;}else memset(num, 0, sizeof(bignum_t)), num[0] = 1, sgn = 0;}else if (!sgn)add(num, a.num), sgn = -a.sgn; return *this;}inline bignum& operator-=(const int a) {if (sgn * a < 0)add(num, ABS(a)); else if (sgn && a) {int ret = comp(num, ABS(a)); if (ret > 0)sub(num, ABS(a)); else if (ret < 0) {bignum_t t;memcpy(t, num, sizeof(bignum_t)); memset(num, 0, sizeof(bignum_t)); num[0] = 1; add(num, ABS(a)); sub(num, t); sgn = -sgn;}else memset(num, 0, sizeof(bignum_t)), num[0] = 1, sgn = 0;}else if (!sgn)sgn = -SGN(a), add(num, ABS(a)); return *this;}inline bignum operator-(const bignum& a) { bignum ret; memcpy(ret.num, num, sizeof(bignum_t)); ret.sgn = sgn; ret -= a; return ret; }inline bignum operator-(const int a) { bignum ret; memcpy(ret.num, num, sizeof(bignum_t)); ret.sgn = sgn; ret -= a; return ret; }inline bignum& operator*=(const bignum& a) { bignum_t t; mul(t, num, a.num); memcpy(num, t, sizeof(bignum_t)); sgn *= a.sgn; return *this; }inline bignum& operator*=(const int a) { mul(num, ABS(a)); sgn *= SGN(a); return *this; }inline bignum operator*(const bignum& a) { bignum ret; mul(ret.num, num, a.num); ret.sgn = sgn * a.sgn; return ret; }inline bignum operator*(const int a) { bignum ret; memcpy(ret.num, num, sizeof(bignum_t)); mul(ret.num, ABS(a)); ret.sgn = sgn * SGN(a); return ret; }inline bignum& operator/=(const bignum& a) { bignum_t t; div(t, num, a.num); memcpy(num, t, sizeof(bignum_t)); sgn = (num[0] == 1 && !num[1]) ? 0 : sgn * a.sgn; return *this; }inline bignum& operator/=(const int a) { int t; div(num, ABS(a), t); sgn = (num[0] == 1 && !num[1]) ? 0 : sgn * SGN(a); return *this; }inline bignum operator/(const bignum& a) { bignum ret; bignum_t t; memcpy(t, num, sizeof(bignum_t)); div(ret.num, t, a.num); ret.sgn = (ret.num[0] == 1 && !ret.num[1]) ? 0 : sgn * a.sgn; return ret; }inline bignum operator/(const int a) { bignum ret; int t; memcpy(ret.num, num, sizeof(bignum_t)); div(ret.num, ABS(a), t); ret.sgn = (ret.num[0] == 1 && !ret.num[1]) ? 0 : sgn * SGN(a); return ret; }inline bignum& operator%=(const bignum& a) { bignum_t t; div(t, num, a.num); if (num[0] == 1 && !num[1])sgn = 0; return *this; }inline int operator%=(const int a) { int t; div(num, ABS(a), t); memset(num, 0, sizeof(bignum_t)); num[0] = 1; add(num, t); return t; }inline bignum operator%(const bignum& a) { bignum ret; bignum_t t; memcpy(ret.num, num, sizeof(bignum_t)); div(t, ret.num, a.num); ret.sgn = (ret.num[0] == 1 && !ret.num[1]) ? 0 : sgn; return ret; }inline int operator%(const int a) { bignum ret; int t; memcpy(ret.num, num, sizeof(bignum_t)); div(ret.num, ABS(a), t); memset(ret.num, 0, sizeof(bignum_t)); ret.num[0] = 1; add(ret.num, t); return t; }inline bignum& operator++() { *this += 1; return *this; }inline bignum& operator--() { *this -= 1; return *this; };inline int operator>(const bignum& a) { return sgn > 0 ? (a.sgn > 0 ? comp(num, a.num) > 0:1) : (sgn < 0 ? (a.sgn < 0 ? comp(num, a.num) < 0 : 0) : a.sgn < 0); }inline int operator>(const int a) { return sgn > 0 ? (a > 0 ? comp(num, a) > 0:1) : (sgn < 0 ? (a < 0 ? comp(num, -a) < 0 : 0) : a < 0); }inline int operator>=(const bignum& a) { return sgn > 0 ? (a.sgn > 0 ? comp(num, a.num) >= 0 : 1) : (sgn < 0 ? (a.sgn < 0 ? comp(num, a.num) <= 0 : 0) : a.sgn <= 0); }inline int operator>=(const int a) { return sgn > 0 ? (a > 0 ? comp(num, a) >= 0 : 1) : (sgn < 0 ? (a < 0 ? comp(num, -a) <= 0 : 0) : a <= 0); }inline int operator<(const bignum& a) { return sgn < 0 ? (a.sgn < 0 ? comp(num, a.num)>0:1) : (sgn > 0 ? (a.sgn > 0 ? comp(num, a.num) < 0 : 0) : a.sgn > 0); }inline int operator<(const int a) { return sgn < 0 ? (a < 0 ? comp(num, -a)>0:1) : (sgn > 0 ? (a > 0 ? comp(num, a) < 0 : 0) : a > 0); }inline int operator<=(const bignum& a) { return sgn < 0 ? (a.sgn < 0 ? comp(num, a.num) >= 0 : 1) : (sgn > 0 ? (a.sgn > 0 ? comp(num, a.num) <= 0 : 0) : a.sgn >= 0); }inline int operator<=(const int a) { return sgn < 0 ? (a < 0 ? comp(num, -a) >= 0 : 1) : (sgn > 0 ? (a > 0 ? comp(num, a) <= 0 : 0) : a >= 0); }inline int operator==(const bignum& a) { return (sgn == a.sgn) ? !comp(num, a.num) : 0; }inline int operator==(const int a) { return (sgn * a >= 0) ? !comp(num, ABS(a)) : 0; }inline int operator!=(const bignum& a) { return (sgn == a.sgn) ? comp(num, a.num) : 1; }inline int operator!=(const int a) { return (sgn * a >= 0) ? comp(num, ABS(a)) : 1; }inline int operator[](const int a) { return digit(num, a); }friend inline istream& operator>>(istream& is, bignum& a) { read(a.num, a.sgn, is); return is; }friend inline ostream& operator<<(ostream& os, const bignum& a) { if (a.sgn < 0)os << '-'; write(a.num, os); return os; }friend inline bignum sqrt(const bignum& a) { bignum ret; bignum_t t; memcpy(t, a.num, sizeof(bignum_t)); sqrt(ret.num, t); ret.sgn = ret.num[0] != 1 || ret.num[1]; return ret; }friend inline bignum sqrt(const bignum& a, bignum& b) { bignum ret; memcpy(b.num, a.num, sizeof(bignum_t)); sqrt(ret.num, b.num); ret.sgn = ret.num[0] != 1 || ret.num[1]; b.sgn = b.num[0] != 1 || ret.num[1]; return ret; }inline int length() { return ::length(num); }inline int zeronum() { return ::zeronum(num); }inline bignum C(const int m, const int n) { combination(num, m, n); sgn = 1; return *this; }inline bignum P(const int m, const int n) { permutation(num, m, n); sgn = 1; return *this; }
};
//以上为高精度模板
typedef bignum ll;ll x1[10001], x2[10001], x3[10001], tot[10001];//x1为1°烷基个数，x2为2°烷基个数，x3为3°烷基个数，tot为烷基个数之和，序数为碳数
void f_calc_alkyl(int n) {x1[1] = 1; tot[1] = 1;x1[2] = 1; tot[2] = 1;x1[3] = 1; x2[3] = 1; tot[3] = 2;x1[4] = 2; x2[4] = 1; x3[4] = 1; tot[4] = 4;//初始化1到4碳的烷基种类 for (int i = 5; i <= n; i++){x1[i] = tot[i - 1];//计算1°烷基种类 for (int j = 1; j < (i - 1) / 2; j++){x2[i] += tot[j] * tot[i - 1 - j];}if (i % 2)x2[i] += tot[(i - 1) / 2] * (tot[(i - 1) / 2] + 1) / 2;elsex2[i] += tot[i / 2] * tot[i / 2 - 1];//计算2°烷基种类 for (int j = 1; j < i; j++){for (int k = j + 1; k < i; k++){if (i - 1 - j - k > k){x3[i] += tot[j] * tot[k] * tot[i - 1 - j - k];}}}//连接的3个烷基为ABC型 for (int j = 1; j < i && i - 1 - 2 * j>0; j++){if (j != i - 1 - 2 * j)x3[i] += (tot[j] * (tot[j] + 1)) / 2 * tot[i - 1 - 2 * j];}//连接的3个烷基为AAB型 if ((i - 1) % 3 == 0){int tem = (i - 1) / 3;x3[i] += tot[tem] * (tot[tem] + 1) * (tot[tem] + 2) / 6;}//连接的3个烷基为AAA型 //计算3°烷基种类 tot[i] = x1[i] + x2[i] + x3[i];//求和 }
}
//计算烷基个数
bignum f_calc_3(int n) {bignum ans;ans = 0;int ub = (n + 1) / 2;//所连烷基碳数上限 for (int i = 1; i < ub; i++){for (int j = i + 1; j < ub; j++){if (n - i - j > j && n - i - j < ub)ans = ans + tot[i] * tot[j] * tot[n - i - j];}}//烷基为ABC型 for (int i = 1; i < ub; i++){if (n - 2 * i > 0 && n - 2 * i < ub && i != n - 2 * i)ans = ans + (tot[i] + 1) * tot[i] / 2 * tot[n - 2 * i];}//烷基为AAB型 if (n % 3 == 0)ans = ans + tot[n / 3] * (tot[n / 3] + 1) * (tot[n / 3] + 2) / 6;//烷基为AAA型return ans;
}
// 计算中心为3°碳的烷烃个数
bignum f_calc_4(int n) {bignum ans;ans = 0;int ub = (n + 1) / 2;for (int i = 1; i < ub; i++) {for (int j = i + 1; j < ub; j++){for (int k = j + 1; k < ub; k++){if (n - i - j - k > k && n - i - j - k < ub)ans += tot[i] * tot[j] * tot[k] * tot[n - i - j - k];}}}//烷基为ABCD型 for (int i = 1; i < ub; i++) {for (int j = 1; j < ub; j++){if (n - 2 * i - j > j && i != j && i != n - 2 * i - j && n - 2 * i - j < ub)ans += tot[i] * (tot[i] + 1) / 2 * tot[j] * tot[n - 2 * i - j];}}//烷基为AABC型 if (n % 2 == 0) {int tem = n / 2;for (int i = 1; i < tem - i; i++)ans += tot[i] * (tot[i] + 1) / 2 * tot[tem - i] * (tot[tem - i] + 1) / 2;}//烷基为AABB型 for (int i = 1; i <= n && n - 3 * i > 0; i++) {if (i != n - 3 * i && n - 3 * i < ub)ans += tot[i] * (tot[i] + 1) * (tot[i] + 2) / 6 * tot[n - 3 * i];}//烷基为AAAB型 if (n % 4 == 0) {ans += tot[n / 4] * (tot[n / 4] + 1) * (tot[n / 4] + 2) * (tot[n / 4] + 3) / 24;}//烷基为AAAA型 return ans;
}
// 计算中心为4°碳的烷烃个数
bignum f_calc_odd(int n) {bignum ans = tot[n / 2] * (tot[n / 2] + 1) / 2 + tot[n / 2] * (tot[n / 2 + 1] - tot[n / 2]);ans += f_calc_3(n - 1);ans += f_calc_4(n - 1);return ans;
}
bignum f_calc_even(int n) {bignum ans = (tot[n / 2] + 1) * tot[n / 2] / 2;ans += f_calc_3(n - 1);ans += f_calc_4(n - 1);return ans;
}
bignum f_calc(int n) {f_calc_alkyl(n);if (n % 2 == 0)return f_calc_even(n);elsereturn f_calc_odd(n);//分奇偶讨论
}
int main()
{int num;cout << "请输入C原子的个数  " << endl;cin >> num;cout <<endl <<"能构成同分异构体的数目为" << f_calc(num) << endl;
}

输入：C原子的个数

输出：该C原子数目所能构成链结构烷烃同分异构体的数目

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