摘要: 注意题目的一个关键条件(a-1)2 2 #include 3 #include 4 #include 5 #include 6 #include 7 #include 8 #include 9 #include 10 using namespace std;11 #define INF 0x3fffffff12 13 typedef __int64 LL;14 15 LL a,b,n,m;16 LL g[2][2];17 LL MOD;18 19 void cal(LL s[2][2],LL t[2][2])20 {21 LL tmp[2][2];22 memse... 阅读全文
posted @ 2013-10-03 22:00 chenhuan001 阅读(335) 评论(0) 推荐(0) 编辑
摘要: int stk[N],vis[N],low[N],link[N],mark[N];int top,index,id,du[N];//记录入度数int pre[N],cnt,g[N];// g 用来记录topsort后的结果int g1[N]; //用来记录缩点后的每一个点所含的点void dfs(int s){ mark[s]=1; vis[s]=index++; low[s]=vis[s]; stk[top++]=s; for(int p=pre[s];p!=-1;p=edge[p].next) { int v=edge[p].to; ... 阅读全文
posted @ 2013-10-03 17:14 chenhuan001 阅读(351) 评论(0) 推荐(0) 编辑
摘要: 数位dp每次都给我一种繁琐的感觉。。A -Palindromic NumbersTime Limit:2000MSMemory Limit:32768KB64bit IO Format:%lld & %lluSubmitStatusDescriptionA palindromic number or numeral palindrome is a 'symmetrical' number like 16461 that remains the same when its digits are reversed. In this problem you will be g 阅读全文
posted @ 2013-10-03 12:10 chenhuan001 阅读(327) 评论(1) 推荐(0) 编辑