POJ--3181--Dollar Dayz--揹包/高精度

Dollar Dayz

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4220		Accepted: 1642

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

題意：輸入n，k表示有k種硬幣，價值是1~k，求用這些硬幣組合出價值n的方案數

解析：數值很大，64位裝不下，所以用兩個64位的分別充當高位和低位來解決，然後就是完全揹包問題了

*******這裏有個點我要說，雖然不算錯，但是這是題目的漏洞跟思維的嚴謹，就是當有高位需要輸出時一定要把位補起來，一般都是以1E18爲摸，所以一般都是有高位就輸出高位，接着把低位輸出，我說如果低位沒有18位整的呢？

拿100來說，高位是1，低位是99，這好，輸出來直接就是199，如果低位是9呢？輸出的就是19，低位的前導零沒有處理啊，有些代碼的正確只是建立在數據的不嚴謹啊。。

#include <iostream>
#include <cstdio>
#include <cstring>
#define Max(a,b) a>b?a:b
using namespace std;
__int64 mod=1000000000000000000;
int main (void)
{
    int n,m,i,j,k,l;
    __int64 dp[2][1111];
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        memset(dp,0,sizeof(dp));
        dp[0][0]=1;
        if(m>n)m=n;
        for(i=1;i<=m;i++)
        for(j=i;j<=n;j++)
        {
            dp[1][j]=dp[1][j-i]+dp[1][j]+(dp[0][j-i]+dp[0][j])/mod;	//高位
            dp[0][j]=(dp[0][j-i]+dp[0][j])%mod;	//低位
        }
        if(dp[1][n])
        {
            printf("%I64d",dp[1][n]);
            printf("%018I64d\n",dp[0][n]);	//補充前導零
        }else
        {
            printf("%I64d\n",dp[0][n]);	//這個不用補充前導零
        }
    }
    return 0;
}