Azarath Metrion Zinthos

E. Petya and Posttime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputLittle Vasya's uncle is a postman. The po

--------const int maxn=400;const int maxm=200000;int n,m;char a[maxn][maxn];char b[maxn][maxn];int stk[10];int cnt;char s[maxm];char cd[7][3]={"1","CX","V","BY","2","AZ","H"};char sp[7][8]=

--------int n;int f[maxn];char s[maxn];int main(){ int n=0; while (gets(s+n)){ n=strlen(s); } s[n]=0; f[0]=0; for (int i=1;i<=n;i++) f[i]=(f[i-1]+1999)%i; if (

The More The BetterTime Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2568    Accepted Submission(s): 668Problem DescriptionGiven a

I-numberTime Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 195    Accepted Submission(s): 75Problem DescriptionThe I-number of x i

#include #include using namespace std;const int maxn=11111;const int maxh=10000019;int head[maxh];int next[maxh];long long st[maxn];void hash_init(){ memset(head,0,sizeof(head));}in

算法设计基础①思维的体操②问题求解常见策略General Problem Solving Techniques③高效算法设计举例Designing Efficient Algorithms ④动态规划专题⑤小结与习题暴力NEFU 722 Anagram 全排列 STL枚举UVa 10755

Problem E: Compound WordsYou are to find all the two-word compound words in a dictionary. A two-word compound word is a word in the dictionary that is the concatenation of exactly two other word

Substrings expTime Limit 1000msMemory Limit 65536KdescriptionYou are given a number of case-sensitive strings of alphabetic characters, find t

Garbage HeapTime limit: ? secondsMemory limit: 64 megabytesFarmer John has a heap of garbage formed in a rectangular parallelepiped.It consists of  garbage pieces each of which has a value.

Number GuessingTime Limit 1000msMemory Limit 65536KdescriptionNumber Guessing is a computer game. First, the computer chooses four different d

Paleontologists in Siberia have recently found a number of fragments of Jurassic period dinosaur skeleton. The paleontologists have decided to forward them to the paleontology museum. Unfortunately, t

int gauss(int r, int c){ bool flag=false; int coe=1; int i=0,t=0; for(int j=0; j<c; ++j) { int index=i; for(int k=i; k<r; ++k) if(a[k][j]>0)

Meeting point-2Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 889    Accepted Submission(s): 485Problem DescriptionIt has been

Meeting point-1Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2324    Accepted Submission(s): 735Problem DescriptionIt has been

二分查找的基本实现int b_search(int key,int a[],int n){ int l=0,r=n-1; while (l<=r) { int m=(l+r)/2; if (a[m]==key) return m; if (a[m]<key) l=m+1; else r=m-1;

E. Subsegmentstime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputProgrammer Sasha has recently begun to study

----------------int n,m;int w[maxn],b[maxn];int s1[maxn],s2[maxn];int w1,b1,w2,b2;double ans,mo;bool flag;void work(int x,int y) { w1=w1-w[s1[x]]+w[s2[y]]; w2=w2-w[s2[y]]+w[s1[x]];