huadong_fcy56的博客

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基础： HDU2112  HDU1874  HDU1869 HDU2066加深对k的理解： HDU1599（最小环）  ZOJ3710待续

吃糖果Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 41705    Accepted Submission(s): 12022Problem DescriptionHOHO，终于从Speakless手上赢走了所有

HDU1290

大数MOD#include#include#include#include#include#include#include using namespace std;string s;long long Mod;int main(){ int i; while(cin>>s>>Mod){ long long ans=0,p=1; int L=s.leng

传送门：Tempter of the Bone             大意是给一个矩阵，叫你是否可以在给定的可走路径上不重复地走，在最后一秒走到终点。我用了两个剪枝，且称其为简直001和剪枝002，事实证明001不要都可以，002不要也能过--||。就当练习一下剪枝。特别是002很有用。                                      if( d %

动态规划+组合数+大数#include#include#include#include#include#include#include using namespace std;long long f[60][60],ans[60];long long c[60][60];int i,j;void _getc(){ for(i=0;i<=50;i++) c[i][0]=

顺便练习下STL和二分HDU1248#include#include#include#include#include#include#include #include#includeusing namespace std;setq;bool M[10010];int a[200],L=0;int main(){ int i,j,T,n; q.insert

Holding Bin-Laden Captive!Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 23275    Accepted Submission(s): 10358Problem DescriptionW

前边和后边性别不同！！！不然NP了                               Girls and BoysProblem Descriptionthe second year of the university somebody started a study on the romantic relations between the students.

#include#include#include#include#includeusing namespace std;long long s[21],c[21];void _gets(){ s[0]=0;s[1]=0;s[2]=1; for(i=3;i<=20;i++) s[i]=(i-1)*(s[i-1]+s[i-2]);}void _getc(){

整数划分　DP　母函数　递推

HDU1081 最大字段和 压缩数组

假设有一个一维序列，现在可以相邻两两交换，把它变成有序的序列，求最小交换次数，大家都知道求逆序对。如果现在有一个围圈的序列，一样两两相邻交换，此时怎么搞？

敌兵布阵Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 99913    Accepted Submission(s): 42300Problem DescriptionC国的死对头A国这段时间正在进行军事演习，所以

Problem DescriptionThere must be many A + B problems in our HDOJ , now a new one is coming.Give you two hexadecimal integers , your task is to calculate the sum of them,and print it in hexadecimal

2^x mod n = 1Time Limit: 2 Seconds      Memory Limit:65536 KB Give a number n, find the minimum x that satisfies 2^x mod n = 1.InputOne positive integer on each line, the value of n.

第一次看到这题果断放弃，毕竟几何白痴，第二次刷没做的题的时候突然想到这个三角形面积的向量法：S=|x1*y2-x2*y1|  但是此题可能是凹多边形，所以不能加绝对值，可以画个凹四边形看看。HDU2036#include#include#include#include#include#includeusing namespace std;int main(){ int

进制转化hdu1197#include#include#include#include#include#includeusing namespace std;int ans1,ans2,ans3;int _change(int u,int v){ int t=v,num=0; while(u>0){ if(u<v){ num+=u; u=0; }

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62的判断就是倒过来没有26hdu2089#include#include#include#include#includeusing namespace std;int sum[1000010];bool _find(int v){ int pre=0,ins; while(v>0){ ins=v%10; if(ins==4) return false; if(

反证法证明得到 中点 满足题意#include#include#include#include#include#includeusing namespace std;int a[600];int ans;int main(){ int T,n,i,j; cin>>T; while(T--){ ans=0; scanf("%d",&n); for(i=1;i<

一：pow，曾经用递归溢出void _pow(int a,int b){ while(b){ if(b%2&1) { ans*a;b--; ans=ans*pow(pos(a,b/2),2); } else ans=ans*pow(pos(a,b/2),2); }}改后——>void _pow(int

这些题都可以用动规或者背包做HDU1028HDU2079HDU2082HDU1028#include#include#include#includeusing namespace std;int c1[330],c2[330],ans[330];int n,i,j,m,k;void _get(){ for(i=0;i c1[i]=1;c2[i]=0;

strlen能不用就不用#include#include#include#include#include#include#include#includeusing namespace std;char s[100010];int a[30];int main(){ int i,L; while(gets(s)){ memset(a,0,sizeof(a)

做多了年月日，现在基本就能水过了                                           18岁生日Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 33799    Accepted Submissi

新手熟悉一下迪杰斯特拉。。。                                                一个人的旅行Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 40926    Accepted Subm