Coin Change 硬币换算

思路讲解

将硬币个数也纳入状态设计当中就可以了。

N=250;
vector dp(N+5,vector<ll>(N+5,0));
dp[0][0]=1;
vector<ll> coin={1,5,10,25,50};
for(auto &c:coin){
	FOR(i,1,N){
		FOR(j,1,100){
			if(i-c<0) break; 
			dp[i][j]+=dp[i-c][j-1];
		}
	}
}

AC代码

https://oj.dhu.edu.cn/#/user/training/382/problems/458?catId=13&continueTrain=yes

源代码

#include <bits/stdc++.h>
#define all(vec) vec.begin(),vec.end()
#define pb push_back
#define SZ(a) ((int) a.size())
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (int i = (a); i >= (b); --i)
#define debug(var) cerr << #var <<":"<<var<<"\n";
#define lson(var) (var<<1)
#define rson(var) ((var<<1)+1)

using namespace std;

using  ll = long long;using ull = unsigned long long;
using DB=double;using LD=long double;
// using i128 = __int128;
using pdd = pair<DB,DB>;using plb = pair<ll,bool>;
using pll = pair<ll,ll>;
using arr3 = array<ll,3> ;using arr2 = array<ll,2>;
constexpr ll MAXN=static_cast<ll>(1e6)+10,INF=static_cast<ll>(1e17)+9; // 1e18+9也是素数
constexpr ll MAXM=(ll)1e6+10;constexpr ll MAXV=(ll)1e6+10;
constexpr ll mod=static_cast<ll>(1e9)+7;
constexpr double eps=1e-8;const double pi=acos(-1.0);

ll N,M,Q,X,K,T,lT,A[MAXN];
/*
50 分、25 分、10 分、5 分和 1 分
*/
inline void init(){
	
}
int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	N=250;
	vector dp(N+5,vector<ll>(N+5,0));
	dp[0][0]=1;
	vector<ll> coin={1,5,10,25,50};
	for(auto &c:coin){
		FOR(i,1,N){
			FOR(j,1,100){
				if(i-c<0) break; 
				dp[i][j]+=dp[i-c][j-1];
			}
		}
	}
// #ifdef LOCAL
    // FOR(i,1,N){
    	// cerr<<dp[i]<<" ";
    // }
    // cerr<<"\n";
// #endif
	cin>>lT;
	while(lT--){
		cin>>N;
		ll ans=0;
		FOR(i,1,100){
			ans+=dp[N][i];
		}
		cout<<ans<<"\n";
	}
	return 0;
}
/*

*/

心路历程（WA，TLE，MLE……）