分析：直接统计装完每一份需要多少个袋子即可
1 #include "iostream" 2 #include "cstdio" 3 #include "cstring" 4 using namespace std; 5 const int maxn=100000+10; 6 int a[maxn]; 7 int n,k; 8 int main() 9 { 10 while(cin>>n>>k) 11 { 12 int ans=0; 13 for(int i=0;i<n;i++){ 14 scanf("%d",&a[i]); 15 if(a[i]%k==0) 16 ans+=a[i]/k; 17 else 18 ans+=a[i]/k+1; 19 } 20 if(ans%2==0) 21 ans/=2; 22 else 23 ans=ans/2+1; 24 cout<<ans<<endl; 25 26 } 27 return 0; 28 }
分析：非常坑的一道题，需要考虑的边界情况包括|b|>l,b=0,q=0,q=1,q=-1，然后其他的情况直接用map来统计一下就好了qwq
1 #include "iostream" 2 #include "cstdio" 3 #include "cstring" 4 #include "string" 5 #include "cmath" 6 #include "vector" 7 #include "algorithm" 8 #include "map" 9 using namespace std; 10 const int maxn=100000+10; 11 long long a[maxn]; 12 long long b,q,l,m; 13 int main() 14 { 15 while(cin>>b>>q>>l>>m){ 16 map<long long,long long>mp; 17 for(int i=0;i<m;i++){ 18 cin>>a[i]; 19 mp[a[i]]=1; 20 } 21 if(abs(b)>l){ 22 cout<<"0"<<endl; 23 continue; 24 } 25 if(b==0){ 26 if(mp[0]){ 27 cout<<"0"<<endl; 28 }else{ 29 cout<<"inf"<<endl; 30 } 31 continue; 32 } 33 if(q==0){ 34 if(mp[b]){ 35 if(mp[0]){ 36 cout<<"0"<<endl; 37 }else{ 38 cout<<"inf"<<endl; 39 } 40 }else{ 41 if(mp[0]){ 42 cout<<"1"<<endl; 43 }else{ 44 cout<<"inf"<<endl; 45 } 46 } 47 continue; 48 } 49 int cnt=0; 50 if(q==1){ 51 if(mp[b]){ 52 cout<<"0"<<endl; 53 }else{ 54 cout<<"inf"<<endl; 55 } 56 continue; 57 } 58 int ans=0; 59 if(q==-1){ 60 if(!mp[b]) 61 ans++; 62 if(!mp[-b]) 63 ans++; 64 if(ans!=0){ 65 cout<<"inf"<<endl; 66 }else{ 67 cout<<"0"<<endl; 68 } 69 continue; 70 } 71 while(abs(b)<=l){ 72 //if(abs(q)==1) 73 //flag++; 74 if(!mp[b]) 75 cnt++; 76 //if(flag==2) break; 77 b*=q; 78 } 79 cout<<cnt<<endl; 80 } 81 return 0; 82 }
分析:计算出最大的f的值，我们需要注意区分奇偶位的正负关系，然后统计出最大的区间即可
1 #include "iostream" 2 #include "cstdio" 3 #include "cstring" 4 #include "string" 5 #include "cmath" 6 #include "algorithm" 7 using namespace std; 8 const int maxn=100000+10; 9 long long a[maxn]; 10 long long dp1[maxn],dp2[maxn]; 11 long long f1[maxn],f2[maxn]; 12 int n; 13 int main() 14 { 15 while(cin>>n) 16 { 17 memset(dp1,0,sizeof(dp1)); 18 memset(dp2,0,sizeof(dp2)); 19 memset(f1,0,sizeof(f1)); 20 memset(f2,0,sizeof(f2)); 21 for(int i=1;i<=n;i++){ 22 scanf("%lld",&a[i]); 23 if(i==1) 24 dp1[i]=dp2[i]=a[i]; 25 else 26 dp1[i]=dp2[i]=abs(a[i]-a[i-1]); 27 } 28 for(int i=1;i<=n;i++){ 29 if(i%2) 30 dp1[i]*=-1; 31 else 32 dp2[i]*=-1; 33 } 34 long long ans1=0,ans2=0; 35 for(int i=2;i<=n;i++){ 36 f1[i]=max(f1[i-1]+dp1[i],dp1[i]); 37 ans1=max(ans1,f1[i]); 38 } 39 for(int i=2;i<=n;i++){ 40 f2[i]=max(f2[i-1]+dp2[i],dp2[i]); 41 ans2=max(ans2,f2[i]); 42 } 43 cout<<max(ans1,ans2)<<endl; 44 } 45 return 0; 46 }