Dollar Dayz（DP）

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原创最新推荐文章于 2026-06-15 13:49:27 发布 508 阅读 AI 写同款· GEO 优化 ›

Agent 长期记忆三代演进：从 RAG 到 Memory Graph，mem0 / Zep / LangMem / Letta 深度横评文章摘要 2026年，Agent长期记忆技术迎来关键突破，OpenAI、Anthropic等厂商通过扩大上下文窗口和推出记忆工具，推动长期记忆从工程补丁发展为独立技术层。本文通过分析三代记忆架构演进（RAG→结构化Fact→MemoryGraph）和四大方案（mem0/Zep/LangMem/Letta）的8维度横评，揭示以下核心观点：长上下文≠长期记忆：128K上下文如同RAM，仍需硬盘式记忆层筛选和结构化存储，避免“Context Rot”导致的准确率下降。三代架构演进： Gen1（RAG）：向量检阅读详情

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

用1到k的硬币能组成n的方法总数。

简单的dp
dp[i] = dp[i] + dp[i - k];
但是这题爆ll，手动开两个ll，一个存高位，一个存低位就好了。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<queue>
#include<vector>
#include<algorithm>
#include<string>
#include<cmath>
#include<set>
#include<map>
#include<vector>
#include<stack>
#include<utility>
#include<sstream>
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int maxn = 1005;
const ll mod = 1e16;
ll dp[2][1005];
int main()
{
    #ifdef LOCAL
    freopen("C:\\Users\\ΡΡ\\Desktop\\in.txt","r",stdin);
    //freopen("C:\\Users\\ΡΡ\\Desktop\\out.txt","w",stdout);
    #endif // LOCAL
    ll n,k;
    scanf("%lld%lld",&n,&k);
    memset(dp,0,sizeof(dp));
    dp[0][0] = 1;
    for(ll i = 1;i <= k;i++)
    {
        for(ll j = i;j <= n;j++)
        {
            dp[0][j] = dp[0][j] + dp[0][j - i];
            dp[1][j] = dp[1][j] + dp[1][j - i] + dp[0][j]/mod;
            dp[0][j] = dp[0][j]%mod;
        }
    }
    if(dp[1][n])
        printf("%lld",dp[1][n]);
    printf("%lld\n",dp[0][n]);
    return 0;
}