BZOJ 1655 [Usaco2006 Jan] Dollar Dayz 奶牛商店 01背包+高精度

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).

    约翰到奶牛商场里买工具．商场里有K(1≤K≤100).种工具，价格分别为1，2，…，K美元．约翰手里有N(1≤N≤1000)美元，必须花完．那他有多少种购买的组合呢？

Input

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

    仅一行，输入N，K.

Output

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

    不同的购买组合数．

Sample Input

5 3

Sample Output

5

HINT

传送门

f[j]=f[j]+f[j-i]，一个非常明显的方程。

但是此题需要高精度！

作为粗心人群就直接ull提交了，然后光荣WA

发现要高精度，

但是高精度预估了30位数……

发现极限数据非常长！！却直接submit了！！

吐血……要开40位才够。。

下次代码慢点打。。

#include<bits/stdc++.h>using namespace std;int f[1005][50];void Plus(int x,int y){int len=max(f[x][0],f[y][0]);for (int i=1;i<=len;i++) f[x][i]=f[x][i]+f[y][i];int i=1;while (i<=len){f[x][i+1]+=f[x][i]/10;f[x][i]%=10;if (f[x][len+1]) len++;i++;}f[x][0]=len;}int main(){int n,K;cin>>n>>K;f[0][0]=f[0][1]=1;for (int i=1;i<=K;i++)for (int j=i;j<=n;j++) Plus(j,j-i);for (int i=f[n][0];i;i--) printf("%d",f[n][i]);putchar('\n');return 0;}