Dollar Dayz S(DP+高精)
题目链接:https://www.luogu.com.cn/problem/P6205
有K种工具,由题目的意思可以看出每种的购买次数工具不限,也就是一个完全背包求方案数的问题
但是如果只是一个简单的完全背包代码放上去会发现:
附上代码:
#include<cstdio>
#include<cmath>
#include<ctime>
#include<cstring>
#include<iostream>
#include<map>
#include<set>
#include<stack>
#include<queue>
#include<string>
#include<vector>
#define ll long long
#define ull unsigned long long
#define up_b upper_bound
#define low_b lower_bound
#define m_p make_pair
#define mem(a) memset(a,0,sizeof(a))
#define IOS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define inf 0x3f3f3f3f
#include<algorithm>
using namespace std;
inline ll read()
{
ll x=0,f=1; char ch=getchar();
while(ch<'0'||ch>'9') {
if(ch=='-') f=-1; ch=getchar(); }
while('0'<=ch&&ch<='9') x=x*10+ch-'0', ch=getchar();
return f*x;
}
const int N = 1e3+5;
int dp[N];
int main()
{
int n,k;
n=read(); k=read();
dp[0]=1;
for(int i=1;i<=k;i++)
for(int j=i;j<=n;j++) dp[j]+=dp[j-i];
cout<<dp[n]<<endl;
return 0;
}
这时候发现并没有这么简单,将极限数据输入会发现答案为负数
然后开个long long?会发现依然WA三个点
只好有高精过了这道题
附上vector高精和数组高精代码
vector实现高精:
#include<cstdio>
#include<cmath>
#include<ctime>
#include<cstring>
#include<iostream>
#include<map>
#include<set>
#include<stack>
#include<queue>
#include<string>
#include<vector>
#define ll long long
#define ull unsigned long long
#define up_b upper_bound
#define low_b lower_bound
#define m_p make_pair
#define mem(a) memset(a,0,sizeof(a))
#define IOS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define inf 0x3f3f3f3f
#include<algorithm>
using namespace std;
inline ll read()
{
ll x=0,f=1; char ch=getchar();
while(ch<'0'||ch>'9') {
if(ch=='-') f=-1; ch=getchar(); }
while('0'<=ch&&ch<='9') x=x*10+ch-'0', ch=getchar();
return f*x;
}
vector<int> add(vector<int> &A,vector<int> &B)
{
int la=A.size(),lb=B.size();
vector<int> C;
int temp=0;
for(int i=0;i<la||i<lb;i++)
{
if(i<la) temp+=A[i];
if(i<lb) temp+=B[i];
C.push_back(temp%10);
temp/=10;
}
if(temp) C.push_back(temp);
return C;
}
const int N = 1e3+5;
vector<int> dp[N];
int main()
{
int n,k;
n=read(); k=read();
dp[0].push_back(1);
for(int i=1;i<=k;i++)
for(int j=i;j<=n;j++) dp[j]=add(dp[j],dp[j-i]);
for(int i=dp[n].size()-1;~i;i--) cout<<dp[n][i];
return 0;
}
数组实现高精:
#include<cstdio>
#include<cmath>
#include<ctime>
#include<cstring>
#include<iostream>
#include<map>
#include<set>
#include<stack>
#include<queue>
#include<string>
#include<vector>
#define ll long long
#define ull unsigned long long
#define up_b upper_bound
#define low_b lower_bound
#define m_p make_pair
#define mem(a) memset(a,0,sizeof(a))
#define IOS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define inf 0x3f3f3f3f
#include<algorithm>
using namespace std;
inline ll read()
{
ll x=0,f=1; char ch=getchar();
while(ch<'0'||ch>'9') {
if(ch=='-') f=-1; ch=getchar(); }
while('0'<=ch&&ch<='9') x=x*10+ch-'0', ch=getchar();
return f*x;
}
const int N = 1e3+5;
int dp[N][N];
void add(int a[],int b[])
{
a[0]=max(a[0],b[0]);
for(int i=1;i<=a[0];i++) a[i]+=b[i];
for(int i=1;i<=a[0];i++) a[i+1]+=a[i]/10, a[i]%=10;
if(a[a[0]+1]) a[0]++;
}
int main()
{
int n,k;
n=read(); k=read();
dp[0][0]=1; dp[0][1]=1;//0位储存长度
for(int i=1;i<=k;i++)
for(int j=i;j<=n;j++)
add(dp[j],dp[j-i]);
for(int i=dp[n][0];i>0;i--) cout<<dp[n][i];
return 0;
}
后来进了题解区才发现还有__int128这一说
代码如下
#include<cstdio>
#include<cmath>
#include<ctime>
#include<cstring>
#include<iostream>
#include<map>
#include<set>
#include<stack>
#include<queue>
#include<string>
#include<vector>
#define ll long long
#define ull unsigned long long
#define up_b upper_bound
#define low_b lower_bound
#define m_p make_pair
#define mem(a) memset(a,0,sizeof(a))
#define IOS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define inf 0x3f3f3f3f
#include<algorithm>
using namespace std;
inline ll read()
{
ll x=0,f=1; char ch=getchar();
while(ch<'0'||ch>'9') {
if(ch=='-') f=-1; ch=getchar(); }
while('0'<=ch&&ch<='9') x=x*10+ch-'0', ch=getchar();
return f*x;
}
//因为cout与printf都不支持直接输出__int128,所以用快写
inline void write(__int128 x)
{
if(x<0) putchar('-'),x=-x;
if(x>9) write(x/10);
putchar(x%10+'0');
}
typedef __int128 bigll;
const int N = 1e3+5;
bigll dp[N]={
1};
int main()
{
int n,k;
n=read(); k=read();
for(int i=1;i<=k;i++)
for(int j=i;j<=n;j++) dp[j]+=dp[j-i];
write(dp[n]);
return 0;
}