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POJ - 2785 4 Values whose Sum is 0
Time Limit: 15000MS Memory Limit: 228000K Total Submissions: 28911 Accepted: 8744 Case Time Limit: 5000MS DescriptionThe SUM problem can be formulated as follows: given four li...原创 2018-08-06 10:18:48 · 227 阅读 · 0 评论 -
HDU - 6383 p1m2 （二分答案）
p1m2Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 1203 Accepted Submission(s): 457Problem Description度度熊很喜欢数组！！我们称一个整数数组为稳定的，若且唯若...原创 2018-08-22 14:45:37 · 279 阅读 · 0 评论 -
POJ - 1064 Cable master （二分）
Cable masterTime Limit: 1000MS Memory Limit: 10000K Total Submissions: 65404 Accepted: 13465 DescriptionInhabitants of the Wonderland have decided to hold a regional programmin...原创 2018-08-22 18:23:00 · 153 阅读 · 0 评论 -
HDU - 3763 CD『二分』
CDTime Limit: 12000/6000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1288 Accepted Submission(s): 558Problem DescriptionJack and Jill have decided to se...原创 2018-08-27 11:05:51 · 158 阅读 · 0 评论 -
二分查找（模板）
int binary(){ int l=0，r=maxn; while(l<=r){ int mid=(l+r)/2; if(find(mid)){ ans=mid; l=mid+1; }else{ r=mid-1; } } cout<<ans<<endl; }可以解决的问题： 求最大的最小值（NOIP2015跳...原创 2018-08-20 10:55:50 · 1204 阅读 · 2 评论