随笔分类 - HDU
摘要：View Code 1 #include<stdio.h> 2 #include<string.h> 3 #define int64 __int64 4 #define N 110 5 int64 dp[N][N],f[N]; 6 int main(){ 7 int64 n,i,j,k,p; 8 while(scanf("%I64d%I64d",&n,&k)!=EOF,n+k){ 9 memset(dp,0,sizeof(dp));10 for(i=1;i<=n;i++) {11 scanf("%I64d",
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摘要：#include<stdio.h>#include<stdlib.h>int cmp( const void *a , const void *b ) { return *(double *)a > *(double *)b ? 1 : -1; }int main(){ int t,i,j,n,v,ansv,tpcnt; double s,ansp,tp; double a[110]; scanf("%d",&t); while(t--){ scanf("%d%d%lf",&n,&v,&s);
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摘要：View Code 1 #include<cstdio> 2 #include<cstdlib> 3 #include<cstring> 4 #include<iostream> 5 #include<queue> 6 #include<algorithm> 7 #define N 55 8 int map[N][N][N]; 9 int a,b,c,maxt,ans;10 int fmin(int i,int j){11 return i>j?j:i;12 }13 using namespace std;14 st
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摘要：#include<stdio.h>#include<string.h>#include<stdlib.h>struct tree{ int lev; struct tree*next[26];};tree root;char str[12];void creat_tree(char *str){ int i,j,len,id; len=strlen(str); tree *p=&root,*temp; for(i=0;i<len;i++){ id=str[i]-'a'; if(p->next[id]==NULL){ tem
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摘要：匈牙利算法。。。我套模版。。。。还是不理解。。革命尚未成功，同志尚需努力。。。#include<stdio.h>#include<string.h>#define MAX 505bool map[MAX][MAX];//表示男女生是否相连bool vis[MAX];//表示是否访问过y里的男生int link[MAX];//link[j]表示y中的j元素当前与x中连接的元素int m,n;bool searchPath(int u){ int v; for(v=1;v<=n;v++){ if(map[u][v]&&vis[v]==false){ v
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摘要：不是很理解啊。。。。有点抄袭的成分。。。唉。。。#include<stdio.h>#include<string.h>//#define N 1000#define MAXN 100010int c[MAXN],a[MAXN];int lowbit(int x)//计算lowbit{ return x&(-x);}void add(int i,int val)//将第i个元素更改为val{ while(i<=MAXN) { c[i]+=val; i+=lowbit(i); }}int sum(int i)//求前i项和{ int s=0...
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摘要：#include<stdio.h>#define N 10000000+5int father[N],sum[N];int ans;int find_father(int x){ if(father[x]==x){ return x; } else{ father[x]=find_father(father[x]); } return father[x];}int fmax(int i,int j){ return i>j?i:j;}void u(int x,int y){ int fax,fay; fax=find_father(x); fay=find_father(y)
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摘要：#include<stdio.h>#include<string.h>#define MAX 9999999#define N 205int map[N][N];int vis[N];int d[N];int start,end,n,m;//n个村庄，m条路径void init(){ memset(map,MAX,sizeof(map)); memset(vis,0,sizeof(vis)); memset(d,MAX,sizeof(d));}void dijkstra(int s,int e){ int i,k,j,min; vis[s]=1; for(i=0;i&l
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摘要：#include<stdio.h>#include<string.h>#define N 100+5#define MAX 9999int map[N][N];int d[N];int n,ans;void prim(){ int i,j,k,min; for(i=1;i<=n;i++) d[i]=map[1][i]; ans+=d[1]; d[1]=0;//进入最小生成树 for(i=1;i<=n;i++){ min=MAX; for(j=1;j<=n;j++){ if(min>d[j]&&d[j]!=0){ k=j; min=
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摘要：#include<stdio.h>#include<string.h>#define N 100000+5int step,mod,seed;int flag;int f[N];int main(){ int i,j,k; while(scanf("%d%d",&step,&mod)==2){ for(i=0;i<=mod-1;i++) f[i]=0; seed=0; do { seed=(seed+step)%mod; f[seed]=1; }while(seed!=0); flag=0; for(i=0;i<=mod-
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摘要：View Code #include<stdio.h>int main(){ int n,m,k,t,left; scanf("%d",&t); while(t--) { scanf("%d%d",&n,&m); k=m+1; left=n%k; if(left>0) printf("first\n"); else if(left==0) printf("second\n"); } return 0;}知道一点就好：...
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摘要：1 #include<stdio.h> 2 #include<stdlib.h> 3 struct gg 4 { 5 int j; 6 int f; 7 double p; 8 }gl[1005],tp; 9 int cmp(const void *a,const void *b)10 {11 return (*(struct gg*)b).p>(*(struct gg*)a).p?1:-1;12 }13 int main()14 {15 int m,n,i;16 double left,sum;17 while(scanf("%d%d...
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