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[Leetcode]657. Robot Return to Origin
Easy There is a robot starting at position (0, 0), the origin, on a 2D plane. Given a sequence of its moves, judge if this robot ends up at (0, 0) aft ...
分类:其他好文   时间:2019-08-27 23:26:05    阅读次数:127
Poj1426 DFS模板题
Find The Multiple Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 51502 Accepted: 21541 Special Judge 题目链接:http://poj.org/problem?id=1426 D ...
分类:其他好文   时间:2019-08-25 00:54:37    阅读次数:100
poj 2566 Bound Found
Bound Found Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 7891 Accepted: 2572 Special Judge Description Signals of most probably extra-te ...
分类:其他好文   时间:2019-08-17 19:44:27    阅读次数:69
poj 1293 Duty Free Shop
Duty Free Shop Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 1896 Accepted: 730 Special Judge Description Pedro travelled to Europe to ta ...
分类:其他好文   时间:2019-08-14 20:01:11    阅读次数:82
【Leetcode_easy】997. Find the Town Judge
problem 997. Find the Town Judge 参考 1. Leetcode_easy_997. Find the Town Judge; 完 ...
分类:其他好文   时间:2019-08-07 19:28:26    阅读次数:81
PAT B1021 个位数统计 (15)
AC代码 include include include using namespace std; const int max_n = 1010; char ans[max_n]; int count[10][2] = {0}; int main() { ifdef ONLINE_JUDGE els ...
分类:其他好文   时间:2019-08-06 01:09:03    阅读次数:79
(Easy) Robot Return to Origin LeetCode
Description here is a robot starting at position (0, 0), the origin, on a 2D plane. Given a sequence of its moves, judge if this robot ends up at (0, ...
分类:其他好文   时间:2019-08-05 19:06:21    阅读次数:95
POJ 3190 Stall Reservations
Stall Reservations Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 15069 Accepted: 5270 Special Judge Description Oh those picky N (1 <= N ...
分类:其他好文   时间:2019-08-04 16:48:17    阅读次数:72
POJ 3190 Stall Reservations
Stall Reservations Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 15069 Accepted: 5270 Special Judge Description Oh those picky N (1 <= N ...
分类:其他好文   时间:2019-08-04 16:41:21    阅读次数:84
Leetcode-5150 Longest Chunked Palindrome Decomposition(段式回文)
1 #define _for(i,a,b) for(int i = (a);i < b;i ++) 2 3 bool judge(string s,int i,int j,int len) 4 { 5 _for(k,0,len) 6 if(s[i]!=s[j]) 7 return false; 8 ... ...
分类:其他好文   时间:2019-08-04 13:41:44    阅读次数:82
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