hdu2085-2086

Posted 2020-09-05 奚政

hdu2085

模拟

 1 #include<stdio.h>
 2 long long a[35][2];
 3 void fun(){
 4     a[0][0]=1;
 5     a[0][1]=0;
 6     for(int i=1;i<=33;i++){
 7         a[i][0]=3*a[i-1][0]+2*a[i-1][1];
 8         a[i][1]=a[i-1][0]+a[i-1][1];
 9     }
10     return;
11 }
12 
13 
14 int main(){
15     fun();
16     int N;
17     while(scanf("%d",&N)!=EOF&&N!=-1){
18         printf("%I64d, %I64d\\n",a[N][0],a[N][1]);
19     }
20     return 0;
21 }

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hdu2086

模拟

 1 #include<stdio.h>
 2 
 3 int main(){
 4     long long n;
 5     double ans,a0,ax,c;
 6     while(scanf("%I64d",&n)!=EOF){
 7         scanf("%lf%lf",&a0,&ax);
 8         ans=n*a0+ax;
 9         for(long long i=1;i<=n;i++) {
10             scanf("%lf",&c);
11             ans-=2*(n+1-i)*c;
12         }
13         printf("%.2lf\\n",ans/(n+1));
14     }
15     return 0;
16 }

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