Dollar Dayz POJ

Dollar Dayz

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 6839 Accepted: 2563

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. Farmer John has exactly $5 to spend. He can buy 5 tools at $1 each or 1 tool at $3 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

Source

USACO 2006 January Silver

这些题全是套路。这一题就是一个简单的完全背包问题，但是发现范围很大，因此想到了第一种做法，一言不和就高精度。

#include <iostream>#include <cstdio>using namespace std;const int maxn = 1e3 + 100;string dp[maxn][maxn];string add(string s1,string s2)//s1=s1+s2{    int i,j,sum=0;    int l1 = s1.length();    int l2 = s2.length();    if(l1<l2) {        swap(s1,s2);        swap(l1, l2);    }    for(i=l1-1,j=l2-1; i>=0; i--,j--) {        sum+=s1[i]-'0';        if(j>=0) {            sum+=s2[j]-'0';        }        s1[i]=sum%10+'0';        sum/=10;    }    if(sum) {        s1='1'+s1;    }    return s1;}int main(){    int n, k;    cin >> n >> k;    for(int i = 0; i <= k; ++i) {        dp[i][0] = "1";    }    for(int i = 1; i <= k; ++i) {        for(int j = 0; j <= n; ++j) {            if(j >= i) {                dp[i][j] = add(dp[i - 1][j], dp[i][j - i]);            } else {                dp[i][j] = dp[i - 1][j];            }        }    }    cout << dp[k][n] << endl;    return 0;}

然而超时了。看了别人的思路发现，开两个数组一个存前14位，一个存后14位。

#include <iostream>#include <cstdio>using namespace std;const int maxn = 1e3 + 100;typedef unsigned long long ull;ull dp[maxn][maxn][2];const ull limit = 1e14;int main(){    int n, k;    cin >> n >> k;    for(int i = 0; i <= k; ++i) {        dp[i][0][1] = 1;    }    for(int i = 1; i <= k; ++i) {        for(int j = 0; j <= n; ++j) {            if(j >= i) {                dp[i][j][1] = dp[i - 1][j][1] + dp[i][j - i][1];                dp[i][j][0] = dp[i - 1][j][0] + dp[i][j - i][0];                dp[i][j][0] += dp[i][j][1] / limit;                dp[i][j][1] =  dp[i][j][1] % limit;            } else {                dp[i][j][1] = dp[i - 1][j][1];                dp[i][j][0] = dp[i - 1][j][0];            }        }    }    if(dp[k][n][0]) {        cout << dp[k][n][0];    }    cout << dp[k][n][1] << endl;    return 0;}