摘要： 除了驱动的效果。 阅读全文

摘要： 阅读全文

摘要： 阅读全文

摘要： A Famous Stone CollectorTime Limit: 30000/15000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 793Accepted Submission(s): 292Problem DescriptionMr. B loves to play with colorful stones. There are n colors of stones in his collection. Two stones with the same color are 阅读全文

摘要： 排列Time Limit:1000MSMemory Limit:30000KTotal Submissions:15173Accepted:6148Description题目描述：大家知道，给出正整数n，则1到n这n个数可以构成n！种排列，把这些排列按照从小到大的顺序（字典顺序）列出，如n=3时，列出1 2 3，1 3 2，2 1 3，2 3 1，3 1 2，3 2 1六个排列。任务描述：给出某个排列，求出这个排列的下k个排列，如果遇到最后一个排列，则下1排列为第1个排列，即排列1 2 3…n。比如：n = 3，k=2 给出排列2 3 1，则它的下1个排列为3 1 2，下2个排列为3 2 1， 阅读全文

摘要： BombTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 131072/65536 K (Java/Others)Total Submission(s): 5247Accepted Submission(s): 1815Problem DescriptionThe counter-terrorists found a time bomb in the dust. But this time the terrorists improve on the time bomb. The number sequence of the time bom 阅读全文

摘要： String painterTime Limit: 5000/2000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1312Accepted Submission(s): 554Problem DescriptionThere are two strings A and B with equal length. Both strings are made up of lower case letters. Now you have a powerful string painter. 阅读全文

摘要： 胜利大逃亡(续)Time Limit: 4000/2000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4328Accepted Submission(s): 1458Problem DescriptionIgnatius再次被魔王抓走了(搞不懂他咋这么讨魔王喜欢)……这次魔王汲取了上次的教训，把Ignatius关在一个n*m的地牢里，并在地牢的某些地方安装了带锁的门，钥匙藏在地牢另外的某些地方。刚开始Ignatius被关在(sx,sy)的位置，离开地牢的门在(ex,ey)的位置。I 阅读全文

摘要： Doing HomeworkTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4362Accepted Submission(s): 1760Problem DescriptionIgnatius has just come back school from the 30th ACM/ICPC. Now he has a lot of homework to do. Every teacher gives him a deadline of ha 阅读全文

摘要： 1 #include 2 #include 3 #include 4 using namespace std; 5 typedef pair a1; 6 map b; 7 int main() 8 { 9 int i,j,k,n,m,x,y,z;10 pair a;11 while(scanf("%d",&n)>0)12 {13 for(i=1;i<=n;i++)14 {15 scanf("%d%d%d",&x,&y,&z);16 a.first=x;17 a.second=y;18 // b[a]=z;19 阅读全文