一道经典的Dp..01背包
定义dp[i] 为需要构造的数字为i 的所有方法数
一开始的时候是这么想的
for(i = 1; i <= N; ++i){
for(j = 1; j <= V; ++j){
if(i - a[j] > 0){
dp[i] += dp[i - a[j]];
}
}
}
状态存在冗余， 输出的时候答案肯定不对
但只需要改一下两个for循环的顺序即可。
Source Code:
/*
ID: wushuai2
PROG: money
LANG: C++
*/
//#pragma comment(linker, "/STACK:16777216") //for c++ Compiler
#include <stdio.h>
#include <iostream>
#include <fstream>
#include <cstring>
#include <cmath>
#include <stack>
#include <string>
#include <map>
#include <set>
#include <list>
#include <queue>
#include <vector>
#include <algorithm>
#define Max(a,b) (((a) > (b)) ? (a) : (b))
#define Min(a,b) (((a) < (b)) ? (a) : (b))
#define Abs(x) (((x) > 0) ? (x) : (-(x)))
#define MOD 1000000007
#define pi acos(-1.0) using namespace std; typedef long long ll ;
typedef unsigned long long ull ;
typedef unsigned int uint ;
typedef unsigned char uchar ; template<class T> inline void checkmin(T &a,T b){if(a>b) a=b;}
template<class T> inline void checkmax(T &a,T b){if(a<b) a=b;} const double eps = 1e- ;
const int M = ;
const ll P = 10000000097ll ;
const int INF = 0x3f3f3f3f ;
const int MAX_N = ;
const int MAXSIZE = ; ofstream fout ("money.out");
ifstream fin ("money.in"); int V, N, a[];
ll dp[]; int main(){
int i, j, k, l, m, n, t, s, c, w, q, num;
fin >> V >> N;
for(i = ; i <= V; ++i){
fin >> a[i];
}
dp[] = ;
for(i = ; i <= V; ++i){
for(j = a[i]; j <= N; ++j){
if(j - a[i] >= ){
dp[j] += dp[j - a[i]];
}
}
}
/*
for(i = 1; i <= N; ++i){
for(j = 1; j <= V; ++j){
if(i - a[j] > 0){
dp[i] += dp[i - a[j]];
}
}
}
*/
fout << dp[N] << endl; fin.close();
fout.close();
return ;
}