poj 3181 Dollar Dayz(完全背包)

题目链接

Dollar Dayz
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 6082 Accepted: 2297
Description

Farmer John goes to Dollar Days at The Cow Store and discovers an unlimited number of tools on sale. During his first visit, the tools are selling variously for 1,2, and 3.FarmerJohnhasexactly5 to spend. He can buy 5 tools at 1eachor1toolat3 and an additional 1 tool at $2. Of course, there are other combinations for a total of 5 different ways FJ can spend all his money on tools. Here they are:

    1 @ US$3 + 1 @ US$2    1 @ US$3 + 2 @ US$1    1 @ US$2 + 3 @ US$1    2 @ US$2 + 1 @ US$1    5 @ US$1

Write a program than will compute the number of ways FJ can spend N dollars (1 <= N <= 1000) at The Cow Store for tools on sale with a cost of 1..K (1 <= K <= 100).
Input

A single line with two space-separated integers: N and K.
Output

A single line with a single integer that is the number of unique ways FJ can spend his money.
Sample Input

5 3
Sample Output

5

题目大意：
输入n，和k，问将n用1到k这k个数字进行拆分，有多少种拆分方法。例如：

n=5,k=3 则有n=3+2,n=3+1+1,n=2+1+1+1,n=2+2+1,n=1+1+1+1+1这5种拆分方法

解题思路：

这个题目是个比较明显的动态规划，而且是背包问题，可以写出状态转移方程如下：

用a[i][j]表示考虑到用数j进行拼接时数字i的拼接方法，可以得到状态转移方程如下：
当i小于j时a[i][j]=a[i][j-1]
当i大于等于j时a[i][j]=a[i][j-1]+a[i-j][j]
由于结果比较大，要敲大数模版。
代码如下：

#include<iostream>#include<cstdio>#include<algorithm>#include<vector>#include<cstring>using namespace std;const int inf=0x3f3f3f3f;const int maxn = 100;struct bign{  int len, s[maxn];  bign() {    memset(s, 0, sizeof(s));    len = 1;  }  bign(int num) {    *this = num;  }  bign(const char* num) {    *this = num;  }  bign operator = (int num) {    char s[maxn];    sprintf(s, "%d", num);    *this = s;    return *this;  }  bign operator = (const char* num) {    len = strlen(num);    for(int i = 0; i < len; i++) s[i] = num[len-i-1] - '0';    return *this;  }  string str() const {    string res = "";    for(int i = 0; i < len; i++) res = (char)(s[i] + '0') + res;    if(res == "") res = "0";    return res;  }  bign operator + (const bign& b) const{    bign c;    c.len = 0;    for(int i = 0, g = 0; g || i < max(len, b.len); i++) {      int x = g;      if(i < len) x += s[i];      if(i < b.len) x += b.s[i];      c.s[c.len++] = x % 10;      g = x / 10;    }    return c;  }  void clean() {    while(len > 1 && !s[len-1]) len--;  }  bign operator * (const bign& b) {    bign c; c.len = len + b.len;    for(int i = 0; i < len; i++)      for(int j = 0; j < b.len; j++)        c.s[i+j] += s[i] * b.s[j];    for(int i = 0; i < c.len-1; i++){      c.s[i+1] += c.s[i] / 10;      c.s[i] %= 10;    }    c.clean();    return c;  }  bign operator - (const bign& b) {    bign c; c.len = 0;    for(int i = 0, g = 0; i < len; i++) {      int x = s[i] - g;      if(i < b.len) x -= b.s[i];      if(x >= 0) g = 0;      else {        g = 1;        x += 10;      }      c.s[c.len++] = x;    }    c.clean();    return c;  }  bool operator < (const bign& b) const{    if(len != b.len) return len < b.len;    for(int i = len-1; i >= 0; i--)      if(s[i] != b.s[i]) return s[i] < b.s[i];    return false;  }  bool operator > (const bign& b) const{    return b < *this;  }  bool operator <= (const bign& b) {    return !(b > *this);  }  bool operator == (const bign& b) {    return !(b < *this) && !(*this < b);  }  bign operator += (const bign& b) {    *this = *this + b;    return *this;  }}d[1005][105];istream& operator >> (istream &in, bign& x) {  string s;  in >> s;  x = s.c_str();  return in;}ostream& operator << (ostream &out, const bign& x) {  out << x.str();  return out;} int main(){    int n,k;    scanf("%d%d",&n,&k);    for(int i=1;i<=k;i++)    d[0][i]=1;    for(int i=1;i<=n;i++)    {        for(int j=1;j<=k;j++)        {            if(i<j) d[i][j]=d[i][j-1];            else if(i>=j) d[i][j]=d[i][j-1]+d[i-j][j];         }    }    cout << d[n][k] <<endl;}