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Time Limit: 20 Sec Memory Limit: 256 MB
http://acm.hdu.edu.cn/showproblem.php?pid=4493
Input
Output
The output will be a single number, the average of money she earned for the twelve months. It will be rounded to the nearest penny, preceded immediately by a dollar sign without tail zero. There will be no other spaces or characters in the output.
Sample Input
2
100.00
489.12
12454.12
1234.10
823.05
109.20
5.27
1542.25
839.18
83.99
1295.01
1.75
100.00
100.00
100.00
100.00
100.00
100.00
100.00
100.00
100.00
100.00
100.00
100.00
￥1581.42
￥100
题意
输出十二个月的平均数，注意去掉后置0……
题解：
+0.005，这是一个特别屌的方法
代码：
//qscqesze #include <cstdio> #include <cmath> #include <cstring> #include <ctime> #include <iostream> #include <algorithm> #include <set> #include <vector> #include <sstream> #include <queue> #include <typeinfo> #include <fstream> #include <map> #include <stack> typedef long long ll; using namespace std; //freopen("D.in","r",stdin); //freopen("D.out","w",stdout); #define sspeed ios_base::sync_with_stdio(0);cin.tie(0) #define test freopen("test.txt","r",stdin) #define maxn 200001 #define mod 10007 #define eps 1e-9 int Num; char CH[20]; const int inf=0x3f3f3f3f; const ll infll = 0x3f3f3f3f3f3f3f3fLL; inline ll read() { ll x=0,f=1;char ch=getchar(); while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();} while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();} return x*f; } inline void P(int x) { Num=0;if(!x){putchar(‘0‘);puts("");return;} while(x>0)CH[++Num]=x%10,x/=10; while(Num)putchar(CH[Num--]+48); puts(""); } //************************************************************************************** int dp[30][30][30][30]; int a[maxn]; int b[maxn]; int temp; int solve(int l1,int r1,int l2,int r2) { if(dp[l1][r1][l2][r2]!=-1) return dp[l1][r1][l2][r2]; if(l1>r1||l2>r2) dp[l1][r1][l2][r2]=0; int sum=0; int ans=0; if(l1<=r1) sum+=a[r1]-a[l1-1]; if(l2<=r2) sum+=b[r2]-b[l2-1]; if(l1<=r1) { ans=max(ans,sum-solve(l1+1,r1,l2,r2)); ans=max(ans,sum-solve(l1,r1-1,l2,r2)); } if(l2<=r2) { ans=max(ans,sum-solve(l1,r1,l2+1,r2)); ans=max(ans,sum-solve(l1,r1,l2,r2-1)); } return dp[l1][r1][l2][r2]=ans; } int main() { //test; int t=read(); while(t--) { memset(dp,-1,sizeof(dp)); memset(a,0,sizeof(a)); memset(b,0,sizeof(b)); int n=read(); for(int i=1;i<=n;i++) { temp=read(); a[i]+=a[i-1]+temp; } for(int i=1;i<=n;i++) { temp=read(); b[i]+=b[i-1]+temp; } cout<<solve(1,n,1,n)<<endl; } }
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原文地址：http://www.cnblogs.com/qscqesze/p/4547576.html
