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POJ3181 Dollar Dayz 【母函数】+【高精度】
Dollar Dayz
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 4204 | Accepted: 1635 |
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$1Write 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
#include <stdio.h> #include <algorithm> #include <string.h> using namespace std; #define maxn 1002 typedef long long LL; const LL UP = 1e18;//1000000000000000000; struct Node { LL u, v; Node(LL x = 0, LL y = 0) { u = x; v = y; } friend Node operator + (Node a, Node b) { a.u += b.u; a.v += b.v; if(a.v >= UP) { a.v -= UP; ++a.u; } return a; } } c1[maxn], c2[maxn], a, b(0, 1); int main() { int N, K, i, j, k; scanf("%d%d",&N, &K); c1[0] = b; for(i = 1; i <= K; ++i) { for(j = 0; j <= N; ++j) for(k = 0; k + j <= N; k += i) c2[j+k] = c2[j+k] + c1[j]; for(j = 0; j <= N; ++j) { c1[j] = c2[j]; c2[j] = a; } } if(c1[N].u) printf("%lld%017lld\n", c1[N].u, c1[N].v); else printf("%lld\n", c1[N].v); return 0; }
POJ3181 Dollar Dayz 【母函数】+【高精度】
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