随笔分类 -  排序

• hdu 4351 Digital root多次范围查询
  摘要：线段树，虽然没接触过也不太懂，但实用价值好像还蛮大的。View Code /*Sample Input15101 240 331 4 5231 34 51 5Sample OutputCase #1:8 7 6 4 27 4 2 -1 -19 8 7 6 4HintFor the first query, [1,3] has six subintervals: [1,1],[2,2],[3,3],[1,2],[2,3],[1,3]. Interval sums are 101,240,331,341,571,672, correspondence digital roots are 2,6,7 阅读全文
  posted @ 2012-08-20 17:33 afterward 阅读(247) 评论(1) 推荐(0)
• poj 2299 Ultra-QuickSort【树状数组】
  摘要：求排序的数最少的交换次数。只能相邻的交换。归并排序算法：View Code #include<stdio.h>#include<limits.h>const int MAXN = 500000 + 10;const int INF = INT_MAX;long long tot;//tot为逆序数总数。int arr[MAXN];//将b数组中的元素复制到a数组中。void CopyArray(int a[], int b[], int len, int left){ for(int i = 0; i < len; i++) { a[left++] = b[i]; 阅读全文
  posted @ 2012-08-10 12:39 afterward 阅读(245) 评论(0) 推荐(0)
• hdu 4350 card
  摘要：当想出来只是求余的问题的时候就不难了。其实都不想发的。View Code #include<iostream>#include<stdio.h>using namespace std;int a[53],b[53];int main(){ int t,n,l,r,i,j; int s,w; cin>>t; int d=0; while(t--) { for(i=0;i<52;i++) cin>>a[i]; cin>>n>>l>>r; s=r-l+1; //... 阅读全文
  posted @ 2012-08-09 20:02 afterward 阅读(262) 评论(0) 推荐(0)
• hdu1285 确定比赛名次【拓扑排序】
  摘要：关于拓扑排序特意找了个简单的做，呵呵。。。View Code #include <stdio.h>#include <string.h>#include <iostream>using namespace std;int a[551][501],ru[551];int main(){ int n,m,i,q1,q2,z,j,k; while(scanf("%d%d",&n,&m)!=EOF) { z=0; memset(a,0,sizeof(a)); memset(ru,0,sizeof(ru)); for(i=1; i. 阅读全文
  posted @ 2012-08-04 19:38 afterward 阅读(167) 评论(0) 推荐(0)
• poj1007 DNA排序
  摘要：View Code #include <iostream>#include <algorithm>using namespace std;class DNA{public:char DNAString[51];int length;int measure;DNA(){ length=0; measure=0;}DNA(const DNA &a ){ length=a.length; measure=a.measure; for (int i=0;i<length;++i ) DNAString[i]=a.DNAString[i]; DNAString[le 阅读全文
  posted @ 2012-08-02 20:49 afterward 阅读(115) 评论(0) 推荐(0)
• acm hdu4325 Flowers（区间最大叠加数）
  摘要：先展示一个我同学些的代码，超级goodView Code #include<iostream>#include<cstdio>#include<cstring>using namespace std;int C[150005];int B[150005];//int Out[150005];int N=150005;int Lowbit(int x){ return x&(-x);}void Modify(int i,int x){ while(i>0) { cout<<i<<" : "<< 阅读全文
  posted @ 2012-08-01 09:34 afterward 阅读(368) 评论(1) 推荐(0)