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hdu 5008(2014 ACM/ICPC Asia Regional Xi'an Online ) Boring String Problem(后缀数组&二分)
Boring String Problem Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 219    Accepted Submission(s): 45 Problem Description In this pro...
分类:其他好文   时间:2014-09-14 23:44:07    阅读次数:447
2014 ACM/ICPC Asia Regional Xi'an Online(HDU 5007 ~ HDU 5017)
题目链接A题 :(字符串查找,水题)题意 :输入字符串,如果字符串中包含“Apple”, “iPhone”, “iPod”, “iPad” 就输出 “MAI MAI MAI!”,如果出现“Sony”就输出“SONY DAFA IS GOOD!” ,大小写敏感。思路 :字符串查找,水题。 1 #inc...
分类:其他好文   时间:2014-09-14 22:00:08    阅读次数:641
HDU 5009 Paint Pearls _(:зゝ∠)_2014 ACM/ICPC Asia Regional Xi'an Online
呵呵 #include #include #include #include typedef long long ll; using namespace std; const int N = 5 * 10000 + 5; int xval[N], dep; int n, a[N], pre[N]; ll d[N]; int pos[300], dd; void work() { d...
分类:其他好文   时间:2014-09-14 20:49:28    阅读次数:211
HDU 5014 Number Sequence 贪心 2014 ACM/ICPC Asia Regional Xi'an Online
尽可能凑2^x-1 #include #include const int N = 100005; int a[N], p[N]; int init(int x) { int cnt = 0; while(x > 1) { x /= 2; cnt ++; } return cnt + 1; } int main() { int n; while(~scanf("%d", ...
分类:其他好文   时间:2014-09-14 20:49:07    阅读次数:165
2014 ACM/ICPC Asia Regional Anshan Online
默默的签到Osu!http://acm.hdu.edu.cn/showproblem.php?pid=5003 1 #include 2 #include 3 using namespace std; 4 int a[64]; 5 int main(){ 6 int t,n; 7 w...
分类:其他好文   时间:2014-09-14 20:32:27    阅读次数:262
HDU 5000 Clone(离散数学+DP)(2014 ACM/ICPC Asia Regional Anshan Online)
Problem DescriptionAfter eating food from Chernobyl, DRD got a super power: he could clone himself right now! He used this power for several times. He...
分类:其他好文   时间:2014-09-14 12:39:27    阅读次数:299
The 2014 ACM-ICPC Asia Regional Anshan Online
【B】RotateTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/65536 K (Java/Others) Special Judge【Problem Description】Noting is more interesting ...
分类:其他好文   时间:2014-09-14 00:02:46    阅读次数:509
poj 5001 Walk &&2014 ACM/ICPC Asia Regional Anshan Online 1005(dp)
http://acm.hdu.edu.cn/showproblem.php?pid=5001 思路:dp计算出途径每个点的总概率,1-x即为所求解。 dp题,先介绍下dp[i][j]为第j步走在第i个点的概率,那么dp[i][j]=dp[x1][j-1]+dp[x2][j-1]+...,x1,x2为i 的相邻节点。上一步在相邻节点这一步才能走到该点嘛。 每个点概率要一个一个的算,当算到第ii...
分类:其他好文   时间:2014-09-13 21:28:55    阅读次数:193
POJ1860:Currency Exchange(BF)
http://poj.org/problem?id=1860DescriptionSeveral currency exchange points are working in our city. Let us suppose that each point specializes in two p...
分类:其他好文   时间:2014-09-13 20:08:45    阅读次数:410
2014 ACM/ICPC Asia Regional Anshan Online
已经确定了的。。。B Rotate 1 /* 2 ID:esxgx1 3 LANG:C++ 4 PROG:B 5 */ 6 #include 7 #include 8 #include 9 #include 10 #include 11 using namespace std;12 13 st...
分类:其他好文   时间:2014-09-13 20:02:15    阅读次数:193
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