POJ 3181(完全揹包、高精度)
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 9693 | Accepted: 3578 |
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
題意就是很簡單的完全揹包,但是結果特別大。需要用兩個long long 陣列 記錄結果。
陣列 a[] 記錄 高位,陣列b[]記錄低 位 (對 1e18 取模結果)
這樣
a[j] = a[j] + a[j-cost[i]] + (b[j] + b[j-cost[i]])%M
b[j] = (b[j] + b[j-cost[i]])%M => b[j] = (b[j]%M + b[j-cost[i]]%M)%M;
code :
#include<iostream>
#include<cmath>
#include<cstring>
#include<cstdio>
#include<vector>
#include<map>
#include<algorithm>
#include<string>
const unsigned long long M = 1e18;
using namespace std;
typedef long long LL;
int cost[100+5];
LL a[1000+5];
LL b[1000+5];
int n,k;
int main()
{
while(scanf("%d%d",&n,&k) != EOF){
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
memset(cost,0,sizeof(cost));
// for(int i = 0; i < k; ++i){
// cost[i] = i+1;
// }
b[0] = 1;
for(int i = 1; i <= k; ++i){
for(int j = i; j <= n; ++j){
a[j] = a[j] + a[j-i] + (b[j] + b[j-i])/M;//記錄高位
b[j] = (b[j]%M + b[j-i]%M)%M; //記錄低位
}
}
if(a[n])
printf("%I64d",a[n]);
printf("%I64d\n",b[n]);
}
return 0;
}