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分析：把”已经使用过的火柴数i”看成状态，可以得到一个图。从前往后每添加一个数字x，就从状态i转移到i+c[x],其中c[x]代表数字x需要的火柴数。当i=0的时候不允许

使用数字0(最后当n>=6时，给答案单独加上1，代表整数0）。

令d（i)为从结点0到结点i的路径条数，则答案f(n)=d(1)+d(2)+d(3)+...+d(n)（因为火柴不必用完，所以使用的火柴的数目可能是1,2,3,...,n）。

程序实现时，我们可以按照从小到大的顺序用d(i)更新所有的d(i+c[j])（j取遍数字0~9），

这道题思路难度是状态转移，而编码难度在大数加法，好久没写大数加法，又有些生疏了，贴上代码：

#include
   
    #include
    
     #include
     
      #include
      
       #include
       
        using namespace std;#define LL long longconst int MAXN=505;const int N=2005;struct bign{int len,num[MAXN];bign (){    len=0;    memset(num,0,sizeof(num));}bign (int number){    *this=number;}void DelZero(){    while(len&&num[len-1]==0)    {        len--;    }    if(len==0)        num[len++]=0;}void put(){    for(int i=len-1;i>=0;i--)        cout<
        
         >n) { dp[n].put(); cout<
        
       
      
     
    
   

#include 
   
    #include 
    
     #include 
     
      using namespace std;const int MAXN = 505;struct bign {    int len, num[MAXN];    bign () {        len = 0;        memset(num, 0, sizeof(num));    }    bign (int number) {*this = number;}    bign (const char* number) {*this = number;}    void DelZero ();    void Put ();    void operator = (int number);    void operator = (char* number);    bool operator <  (const bign& b) const;    bool operator >  (const bign& b) const { return b < *this; }    bool operator <= (const bign& b) const { return !(b < *this); }    bool operator >= (const bign& b) const { return !(*this < b); }    bool operator != (const bign& b) const { return b < *this || *this < b;}    bool operator == (const bign& b) const { return !(b != *this); }    void operator ++ ();    void operator -- ();    bign operator + (const int& b);    bign operator + (const bign& b);    bign operator - (const int& b);    bign operator - (const bign& b);    bign operator * (const int& b);    bign operator * (const bign& b);    bign operator / (const int& b);    //bign operator / (const bign& b);    int operator % (const int& b);};/*Code*/const int N = 2005;const int c[20] = {6, 2, 5, 5, 4, 5, 6, 3, 7, 6};bign dp[N];int main () {    dp[0] = 1;    dp[1] = 0;    for (int i = 0; i <= 2000; i++) {        for (int j = 0; j < 10; j++) {            if (i == 0 && j == 0)                continue;            if (i + c[j] <= 2000)                dp[i+c[j]] = dp[i+c[j]] + dp[i];        }    }    dp[6] = dp[6] + 1;    for (int i = 2; i <= 2000; i++)        dp[i] = dp[i] + dp[i-1];    int n;    while (scanf("%d", &n) == 1 && n > 0) {        dp[n].Put();        printf("\n");    }    return 0;}void bign::DelZero () {    while (len && num[len-1] == 0)        len--;    if (len == 0) {        num[len++] = 0;    }}void bign::Put () {    for (int i = len-1; i >= 0; i--)         printf("%d", num[i]);}void bign::operator = (char* number) {    len = strlen (number);    for (int i = 0; i < len; i++)        num[i] = number[len-i-1] - '0';    DelZero ();}void bign::operator = (int number) {    len = 0;    while (number) {        num[len++] = number%10;        number /= 10;    }    DelZero ();}bool bign::operator < (const bign& b) const {    if (len != b.len)        return len < b.len;    for (int i = len-1; i >= 0; i--)        if (num[i] != b.num[i])            return num[i] < b.num[i];    return false;}void bign::operator ++ () {    int s = 1;    for (int i = 0; i < len; i++) {        s = s + num[i];        num[i] = s % 10;        s /= 10;        if (!s) break;    }    while (s) {        num[len++] = s%10;        s /= 10;    }}void bign::operator -- () {    if (num[0] == 0 && len == 1) return;    int s = -1;    for (int i = 0; i < len; i++) {        s = s + num[i];        num[i] = (s + 10) % 10;        if (s >= 0) break;    }    DelZero ();}bign bign::operator + (const int& b) {    bign a = b;    return *this + a;}bign bign::operator + (const bign& b) {    int bignSum = 0;    bign ans;    for (int i = 0; i < len || i < b.len; i++) {        if (i < len) bignSum += num[i];        if (i < b.len) bignSum += b.num[i];        ans.num[ans.len++] = bignSum % 10;        bignSum /= 10;    }    while (bignSum) {        ans.num[ans.len++] = bignSum % 10;        bignSum /= 10;    }    return ans;}bign bign::operator - (const int& b) {    bign a = b;    return *this - a;}bign bign::operator - (const bign& b) {    int bignSub = 0;    bign ans;    for (int i = 0; i < len || i < b.len; i++) {        bignSub += num[i];        bignSub -= b.num[i];        ans.num[ans.len++] = (bignSub + 10) % 10;        if (bignSub < 0) bignSub = -1;    }    ans.DelZero ();    return ans;}bign bign::operator * (const int& b) {    int bignSum = 0;    bign ans;    ans.len = len;    for (int i = 0; i < len; i++) {        bignSum += num[i] * b;        ans.num[i] = bignSum % 10;        bignSum /= 10;    }    while (bignSum) {        ans.num[ans.len++] = bignSum % 10;        bignSum /= 10;    }    return ans;}bign bign::operator * (const bign& b) {    bign ans;    ans.len = 0;     for (int i = 0; i < len; i++){          int bignSum = 0;          for (int j = 0; j < b.len; j++){              bignSum += num[i] * b.num[j] + ans.num[i+j];              ans.num[i+j] = bignSum % 10;              bignSum /= 10;        }          ans.len = i + b.len;          while (bignSum){              ans.num[ans.len++] = bignSum % 10;              bignSum /= 10;        }      }      return ans;}bign bign::operator / (const int& b) {    bign ans;    int s = 0;    for (int i = len-1; i >= 0; i--) {        s = s * 10 + num[i];        ans.num[i] = s/b;        s %= b;    }    ans.len = len;    ans.DelZero ();    return ans;}int bign::operator % (const int& b) {    bign ans;    int s = 0;    for (int i = len-1; i >= 0; i--) {        s = s * 10 + num[i];        ans.num[i] = s/b;        s %= b;    }    return s;}
     
    
   

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