hdu 2086 A1 = ?
2015-10-12 14:19
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题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2086
题目大意:只怪数学不行 网上看到的觉得很好。。。。
因为:Ai=(Ai-1+Ai+1)/2 - Ci,
A1=(A0 +A2 )/2 - C1;
A2=(A1 + A3)/2 - C2 , ...
=> A1+A2 = (A0+A2+A1+A3)/2 - (C1+C2)
2[(A1+A2)+(C1+C2)] = A0+A2+A1+A3;
A1+A2 = A0+A3 - 2(C1+C2);
=> A1+A2 = A0+A3 - 2(C1+C2)
同理可得:
A1+A1 = A0+A2 - 2(C1)
A1+A2 = A0+A3 - 2(C1+C2)
A1+A3 = A0+A4 - 2(C1+C2+C3)
A1+A4 = A0+A5 - 2(C1+C2+C3+C4)
...
A1+An = A0+An+1 - 2(C1+C2+...+Cn)
----------------------------------------------------- 左右求和
(n+1)A1+(A2+A3+...+An) = nA0 +(A2+A3+...+An) + An+1 - 2(nC1+(n-1)C2+...+2Cn-1+Cn)
=> (n+1)A1 = nA0 + An+1 - 2(nC1+(n-1)C2+...+2Cn-1+Cn)
=> A1 = [nA0 + An+1 - 2(nC1+(n-1)C2+...+2Cn-1+Cn)]/(n+1)
附上代码
题目大意:只怪数学不行 网上看到的觉得很好。。。。
因为:Ai=(Ai-1+Ai+1)/2 - Ci,
A1=(A0 +A2 )/2 - C1;
A2=(A1 + A3)/2 - C2 , ...
=> A1+A2 = (A0+A2+A1+A3)/2 - (C1+C2)
2[(A1+A2)+(C1+C2)] = A0+A2+A1+A3;
A1+A2 = A0+A3 - 2(C1+C2);
=> A1+A2 = A0+A3 - 2(C1+C2)
同理可得:
A1+A1 = A0+A2 - 2(C1)
A1+A2 = A0+A3 - 2(C1+C2)
A1+A3 = A0+A4 - 2(C1+C2+C3)
A1+A4 = A0+A5 - 2(C1+C2+C3+C4)
...
A1+An = A0+An+1 - 2(C1+C2+...+Cn)
----------------------------------------------------- 左右求和
(n+1)A1+(A2+A3+...+An) = nA0 +(A2+A3+...+An) + An+1 - 2(nC1+(n-1)C2+...+2Cn-1+Cn)
=> (n+1)A1 = nA0 + An+1 - 2(nC1+(n-1)C2+...+2Cn-1+Cn)
=> A1 = [nA0 + An+1 - 2(nC1+(n-1)C2+...+2Cn-1+Cn)]/(n+1)
附上代码
1 #include <iostream> 2 #include <cstdio> 3 4 using namespace std; 5 int main () 6 { 7 int i,j,n; 8 double sum; 9 while (~scanf("%d",&n)) 10 { 11 sum=0; 12 double a[3500],c[3500],d; 13 scanf("%lf%lf",&a[0],&a[n+1]); 14 for (i=1; i<=n; i++) 15 scanf("%lf",&c[i]); 16 a[1]=n*a[0]+a[n+1]; 17 for (i = n, j = 1; i >=1&&j <= n; j++,i--) 18 sum += i*c[j]; 19 a[1] = (a[1] - 2*sum)/(n+1); 20 //sum+=c[i]; 21 //cout<<a[1]<<endl; 22 printf ("%.2lf\n",a[1]); 23 } 24 return 0; 25 }
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