2017 ECL-FINAL J.Straight Master

题目链接：http://codeforces.com/gym/101775/problem/J

思路：序列差分一下，然后用得到的查分序列乱搞就可以了

注意差分序列第一项等于a[i]，之后n-1项为cha[i]=a[i]-a[i-1]，第n+1项为0-a[n]

 #include<bits/stdc++.h>
 using namespace std;

 typedef long long ll;
 const int mod=1e9+;
 const int maxn=2e5+;
 const int inf=0x3f3f3f3f;
 const double eps=1e-;
 const double pi=acos(-1.0);
 #define mem(s,v) memset(s,v,sizeof(s))
 #define pdd pair<double,double>
 #define pii pair<int,int>

 int t,n;
 int a[maxn],cha[maxn];

 int main(){
     scanf("%d",&t);
     for(int k=;k<=t;++k){
         scanf("%d",&n);
         for(int i=;i<=n;++i){
             scanf("%d",&a[i]);
             if(i==) cha[i]=a[i];
             else cha[i]=a[i]-a[i-];
         }
         cha[n+]=-a[n];
         int sum=;
         int flag=;
         if(cha[]<||cha[]<) flag=;
         for(int i=;i<=n+;++i){
 //            cout<<cha[i]<<" ";
             if(cha[i]>=) sum+=cha[i];
             int temp=i+;
             if(temp>n+) break;
             if(cha[temp]<) sum+=cha[temp];
             if(sum<) break;
         }
 //        cout<<endl;
         if(sum) flag=;
         printf("Case #%d: ",k);
         if(flag) printf("Yes\n");
         else printf("No\n");
     }
     return ;
 }