摘要: Common Subsequence Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1159DescriptionA subsequence of a given sequence is the given sequence with some elements (possible none) left out. Given a sequence X = another sequence Z = i... 阅读全文
posted @ 2013-07-19 20:39 15HP_EPM测试4_王晓 阅读(109) 评论(0) 推荐(0) 编辑
摘要: Description把M个同样的苹果放在N个同样的盘子里,允许有的盘子空着不放,问共有多少种不同的分法?(用K表示)5,1,1和1,5,1 是同一种分法。Input第一行是测试数据的数目t(0 n时,必然有盘子里无法放苹果,这种情况,f(m,n)=f(m,m);当n 2 #include 3 #include 4 #include 5 int f(int m,int n) 6 { 7 if(m==0||n==1) return 1; 8 else if(m>num;17 while(num--)18 {19 cin>>m>>n;20 ... 阅读全文
posted @ 2013-07-19 20:10 15HP_EPM测试4_王晓 阅读(185) 评论(0) 推荐(0) 编辑