2018 年 8月 13 日随笔档案 - ffgcc

摘要：解析：初始矩阵（B,A,1) 单位矩阵（1,1,0,2,0,0,K,0,1）其中K可以分成好几段,分段具体看代码#include#define ll long long#define inf 0x3f3f3f3f#define pb push_back... 阅读全文

posted @ 2018-08-13 19:23 ffgcc 阅读(117) 评论(0) 推荐(0)

摘要： #include#define ll long long#define inf 0x3f3f3f3f#define pb push_back#define mod 1000000007#define rep(i,a,b) for(int i=a;i=a;i-... 阅读全文

posted @ 2018-08-13 19:03 ffgcc 阅读(123) 评论(0) 推荐(0)

摘要： Swordsman Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 188 Accep... 阅读全文

posted @ 2018-08-13 19:02 ffgcc 阅读(122) 评论(0) 推荐(0)

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