标签:tom enc follow sheng div ring problems mod RoCE
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 35117 Accepted Submission(s): 12510
1 #include<cstdio> 2 using namespace std; 3 const int maxn=1000006; 4 int a[maxn],dp[maxn]; 5 6 int my_max(int x,int y) 7 { 8 return x>y?x:y; 9 } 10 11 int main() 12 { 13 int n,m; 14 while( ~scanf("%d%d",&m,&n)){ 15 16 for(int i=1;i<=n;i++){ 17 scanf("%d",&a[i]); 18 dp[i]=0; 19 } 20 21 int temp; 22 for(int i=1;i<=m;i++){ 23 temp=0; 24 for(int j=1;j<=i;j++) 25 temp+=a[j]; 26 dp[n]=temp; 27 28 for(int j=i+1;j<=n;j++){ 29 temp=my_max( dp[j-1], temp)+a[j]; 30 dp[j-1]=dp[n]; 31 dp[n]=my_max( dp[n], temp); 32 } 33 } 34 35 printf("%d\n",dp[n]); 36 } 37 return 0; 38 }
https://www.cnblogs.com/dongsheng/archive/2013/05/28/3104629.html
1 #include<cstdio> 2 #include<climits> 3 #include<cstring> 4 #include<iostream> 5 const int maxn=1000006; 6 int m,n; 7 int data[maxn], cur[maxn], pre[maxn]; 8 9 int Maxsum() 10 { 11 int max_sum; 12 for(int i=1;i<=m;i++){ 13 max_sum=INT_MIN; 14 15 for(int j=i;j<=n;j++){ 16 if(cur[j-1]<pre[j-1]) 17 cur[j]=pre[j-1]+data[j]; 18 else 19 cur[j]=cur[j-1]+data[j]; 20 21 pre[j-1]=max_sum; 22 23 if(max_sum<cur[j]) 24 max_sum=cur[j]; 25 } 26 27 } 28 return max_sum; 29 } 30 31 int main() 32 { 33 while( ~scanf("%d%d", &m,&n)){ 34 memset( cur, 0, sizeof cur); 35 memset( pre, 0, sizeof pre); 36 37 for(int i=1;i<=n;i++) 38 scanf("%d",&data[i]); 39 printf("%d\n",Maxsum()); 40 } 41 return 0; 42 }
这个相对好理解。
标签:tom enc follow sheng div ring problems mod RoCE
原文地址:https://www.cnblogs.com/ZQUACM-875180305/p/9077413.html