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Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 64178 Accepted Submission(s): 35350
1 #include<iostream> 2 #include<cstdio> 3 #include<cmath> 4 #include<cstring> 5 #define N 100 6 using namespace std; 7 int a[N]; 8 int main() 9 { 10 int n; 11 while(scanf("%d",&n)!=EOF) 12 { 13 if(n==0) 14 break; 15 int sum=n*5; 16 memset(a,0,sizeof(0)); 17 int i; 18 for(i=1;i<=n;i++) 19 { 20 scanf("%d",&a[i]); 21 } 22 int j; 23 a[0]=0; 24 for(i=0;i<n;i++) 25 { 26 j=i+1; 27 if(a[j]>a[i]) 28 sum=sum+abs(a[j]-a[i])*6; 29 if(a[j]<a[i]) 30 sum=sum+abs(a[j]-a[i])*4; 31 } 32 printf("%d\n",sum); 33 } 34 }
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原文地址:http://www.cnblogs.com/Aa1039510121/p/5831481.html