YogyKwan的远行

摘要： D. Sweets for Everyone!time limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputFor he knew every Who down in Whoville beneath, Was busy now, hanging a mistletoe wreath. "And they're hanging their stockings!" he snarled with a sneer, "T 阅读全文

摘要： A. Paper Work模拟：从左到右扫一遍，每组一旦出现第三个负数就新开一组B. Restoring IPv6模拟：根据“::”分情况讨论C. Movie Critics贪心：连续相同的视为一段，每段两端相同，该段权重+2，否则+1。最后找权重最大的去掉即可D. Building Bridge计算几何：通过相似得到x2上没点在x1上对应的坐标，然后在x1的点组里二分找到该坐标左右位置，对于x2上每点只要计算这两个位置跟它对应即可，边扫边找最小E. Mad Joe模拟：遍历每一层时，设置两个参量表示左右边界，重复的路直接通过左右边界的距离来叠加，复杂度m*n4CROC-MBTU 2012, 阅读全文

摘要： A. Cupboards模拟：两列的最小值相加B. Chilly Willy数论：结尾可以找到循环规律C. Robo-Footballer计算几何：使球达到位置为y1+r，然后此时用三角形关系判断是否撞了y2的柱子D. Sweets for Everyone!模拟：列出一个初始值时，通过贪心思想，有就给，没有就去该点的跳转位置（通过队列预处理得到），最后再反过来扫一遍，找到使得时间最少的方法。通过二分初始值，得到答案。E. Piglet's Birthday神题：研究中，附链接http://codeforces.com/contest/248/problem/E3Codeforces 阅读全文

摘要： Trim the NailsTime Limit:2 Seconds Memory Limit:65536 KBRobert is clipping his fingernails. But the nail clipper is old and the edge of the nail clipper is potholed.The nail clipper's edge isNmillimeters wide. And we useNcharacters('.' or '*') to represent the potholed nail clipp 阅读全文

摘要： D.Japanese Mahjong I模拟：枚举每张牌加入，判断是否能组成4*3+2的模式F.Japanese Mahjong III模拟：直接判断是否七对或十三烂G.Gao The Sequence贪心：转化为差数组上的操作。从后往前扫，若目前节余值t加上剩余的差之和仍然当前扫到的差，则NO。一直没有NO就是YES。I.Search in the Wiki暴力：直接用stl的set_intersection判断交集J.Trim the Nailsbfs：每次用没有用过的方式去剪指甲，结果为0，即指甲剪光了就结束45 jingg.cxy4100284 (1)0105 (1)00265 (2) 阅读全文

摘要： E. Blood Cousins Returntime limit per test3 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputPolycarpus got hold of a family tree. The found tree describes the family relations ofnpeople, numbered from 1 ton. Every person in this tree has at most one direct ancestor. 阅读全文

摘要： A. Buggy Sorting贪心：只要n>2，则令最前两个为2，后面都为1B. Increase and Decrease贪心：若所有元素相加整除元素个数n，则为n，否则为n-1C. Beauty Pageant暴力：用vector存枚举到每个元素的当前所有解，没加入一个新元素，更新vector，知道个数等于kD. Colorful Graph图论：输入边时，将对应点加入，最后枚举一遍颜色，取最大E. Blood Cousins Return树：将节点按深度用vector分别存储，递归时每个节点的头\末位置分别记录，搜索v的k儿子时，利用头、末位置调用lower_bound()函数对 阅读全文

摘要： C. Beauty Pageanttime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputGeneral Payne has a battalion ofnsoldiers. The soldiers' beauty contest is coming up, it will last forkdays. Payne decided that his battalion will participate in the pageant. No 阅读全文

摘要： A. System Administrator模拟：直接计算同类第一项相加是否大于等于总和的一半B. Internet Address字符串：关键在于找"ru"C. Game with Coins贪心：保证堆数为奇数，且最后的两堆一定要先取D. Restoring Table位运算：按32位分别做，生成1的两者都为1E. Mishap in Club模拟：能利用已经存在的人就利用，否则重新加人F. Log Stream Analysis模拟：枚举每个事件，它之后的第num-1个的时间减去该事件时间是否<timeG. Suggested Friends图论：用map和v 阅读全文

摘要： G. Suggested Friendstime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputPolycarpus works as a programmer in a start-up social network. His boss gave his a task to develop a mechanism for determining suggested friends. Polycarpus thought much about th 阅读全文