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The shortest problemTime Limit: 3000/1500 MS (Java/Others)Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 0Accepted Submission(s): 0Prob... 阅读全文
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A. Little KengTime Limit: 2000msMemory Limit: 65536KB64-bit integer IO format: %lld Java class name: Main Submit Status Calculate how many 0s at the ... 阅读全文
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Key SetTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 476Accepted Submission(s): 251Problem Des... 阅读全文
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JudgingTroublesTime Limit: 5000ms, Special Time Limit:12500ms, Memory Limit:65536KBTotal submit users: 84, Accepted users: 61Problem 13352 : No specia... 阅读全文
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例题hduJudgingTroubles#include struct node //必须是结构体{ char ch[20];};node ch1[100005],ch2[100005];int comp(node a,node b){ return strcmp(a.ch,b.ch)... 阅读全文
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分析一下上面那个图看能得出什么规律： 据图可知：c1=a1，c2=a1+a2，c3=a3，c4=a1+a2+a3+a4，c5=a5，c6=a5+a6，c7=a7，c8=a1+a2+a3+a4+a5+a6+a7+a8，c9=a9，c10=a9+a10，c11=a11........c16=a1+a2+... 阅读全文
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//n个t相乘，对mod取余long long ksm(int q,int n){ long long s=1,t=q; while(n>0) { if(n%2!=0) { n=n-1; s=s*t%mod; ... 阅读全文
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#include#includeint next[100];void set_next(char ch[]) //带入子串{ int len,j,i; len=strlen(ch); next[0]=-1; j=-1; i=0; while(i0)//str为母... 阅读全文
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//公约数之和打表#include #include #include using namespace std;int a[200005];int main(){ memset(a,0,sizeof(a)); for(int i=1;i#include #include using na... 阅读全文
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double chaji(double ax,double ay,double bx,double by,double cx,double cy) //两条线段斜率的差，double !!!{ return (by-ay)*(cx-ax)-(cy-ay)*(bx-ax);}void co... 阅读全文