菜园

忘记了桌子可以搬至任意房间，我默认为只能从大到小悲剧································································· #include#includeusing namespace std;int main(){ int n,m,i,boxto,boxfrom,room[201],max; cin

prim算法，用double，我第一次用float wa，看了别人代码改成double就过了。#include#include#define N 110#define max 9999999double g[N][N];void prim(int n){ int i,j,u,flag,mark[N]; double dis[N],cost,min;

<br />#include<stdio.h> int main() { int a,b; while(scanf("%d%d",&a,&b)&&(a!=0||b!=0)) { printf("%d/n",a+b); } }

<br />#include<stdio.h> int main() { int a,i,n,sum; while(scanf("%d",&n)&&n) { sum=0; for(i=0;i<n;i++) { scanf("%d",&a); sum+=a; } printf("%d/n",sum); } }

<br />#include<stdio.h> int main() { int m, a,i,n,sum; scanf("%d",&m); while(m--) { scanf("%d",&n); sum=0; for(i=0;i<n;i++) { scanf("%d",&a); sum+=a; } printf("%d/n",sum); } }

<br />#include<stdio.h> int main() { int a,b; while(scanf("%d%d",&a,&b)!=EOF) { printf("%d/n",a+b); } }

<br />#include<stdio.h><br />int main()<br />{<br /> int a,b,i,n,sum;<br /> while(scanf("%d%d",&a,&b)!=EOF)<br /> {<br />  sum=a+b;<br />  printf("%d/n/n",sum);<br /> }<br />}

<br />#include<stdio.h> int main() { int a,b,n; scanf("%d",&n); while(n--) { scanf("%d%d",&a,&b); printf("%d/n",a+b); } }

<br />#include<stdio.h> int main() { int a,i,n,sum; while(scanf("%d",&n)!=EOF) { sum=0; for(i=0;i<n;i++) { scanf("%d",&a); sum+=a; } printf("%d/n",sum); } }

<br />#include<stdio.h> int main() { int m,a,i,n,sum; scanf("%d",&m); while(m--) { scanf("%d",&n); sum=0; for(i=0;i<n;i++) { scanf("%d",&a); sum+=a; } printf("%d/n",sum); if(m!=0) printf("/n"); } }

<br />#include<stdio.h> int main() { long long sum,i,m; while(scanf("%d",&m)!=EOF) { printf("%ld/n/n",m*(m+1)/2); } }<br />读懂题目是成功的一半

<br />#include<stdio.h> int main() { int i; char str[200]; while(gets(str)) { for(i=0;str[i]!='/0';i++) { if(str[i]>='A'&&str[i]<='Z') str[i]+=32; } for(i=0;str[i]!='/0';i++) { printf("%c",str[i]); } printf("/n")

经典的并查集运用，不解释要提高代码的正确率呀！！！！！！！！！！！！！！！！！！！！！！！！#include#define N 55000int set[N];int find(int a){ int i,j,r; i=a; r=a; while(r!=set[r]) r=set[r]; while(i!=r) { j=set[i];

多次PE 太不细心了#include#includeint main(){ char op1[1002],op2[1002]; int c,i,j,n,m,s1[1002],s2[1002],len1,len2,count; count=1; scanf("%d",&n); m=n; while(m--) { memset(s1,0,1002*si

 读不懂题目第一次读，照着sample写程序，wa第二次，仿佛题意是每次字母变为下一个，变换数次，wa第三次，仿佛字母可以随意换·······被文中的例子骗了 #include#includeusing namespace std;int main(){ int i,len,n[2][26]; char code[102],tru[102];

求素数表应该还有更高效的算法#include#include#include#define NUM 10001using namespace std;int main(){ bool prime[NUM]; int len,lenp,i,j,pri[1300],n,count,sum; len=(int)sqrt(NUM); memset(prime,1

用char数组 wa用string类 ac不解#include//#include#includeusing namespace std;string n;int s[103];int main(){ int i,j,len; memset(s,0,sizeof(int)*103); while(1) { cin>>n; //le

速度敲代码#include#include#define N 1000000using namespace std;bool prime[N];void p(void){ int i,j,len; len=(int)sqrt((float)N); memset(prime,1,N); prime[0]=prime[1]=0; for(i=2;i<N/

#include#include#include#define NUM 1000000#define N NUM/2bool prime[N];void p(void);int main(){ int i,n; p(); while(1) { scanf("%d",&n); if(!n||n%2) return 0; for

注意边界条件#include#include#define N 111int cmp(int s[],int t[]){ int i; for(i=N-1;i>=0;i--) { if(s[i]>t[i]) return 1; else if(s[i]<t[i]) return -1; } return 0;}int ma

恶心的题目1.读不懂的题目，首先我没搞清输入输出格式，不知道怎么做2.没注意边界条件 这是非常恶心的，搜索了别人的解题报告才做出这道题因为这道题丧失了A题的斗志，以至于两个星期都没有再碰onlineJudge要提高自己的抗挫折能力 #include#include#define N 110int main(){ int i,j,k,len,n,op[N]

注意边界条件 0#include#include#define N 440void print(int t[]){ int i,j,k,high,low; for(low=0;low<2*N;low++) if(t[low]) break; for(high=2*N-1;high>=0;high--) if(t[high]) break;

注意边界条件，如000010之类的例外，没有. pku1001要比hdu1063弱大数真烦人，谁让我不会java呢 #include#include#define N 200int main(){ int i,j,k,flag,low,high,n,r[N],t[N],result[N]; char input[10]; while(scanf("%s%

#include#include#define N 500int main(){ int i,j,n,p,f2[N],f1[N],f0[N]; scanf("%d",&n); while(n--) { scanf("%d",&p); memset(f0,0,sizeof(f0)); memset(f1,0,sizeof(f1)); mem

第一遍总是做不出来！！！！！！！这次第一次提交tle，自己发现不了错误，网上搜别人的代码才发现问题：我是对每一对（起点，终点）均用一次dijkstra，其实只要对每个起点用一次就可以了，全部终点的信息都包含在里面了。一次dijkstra生成起点到剩余点的所有最值。我对dijkstra算法还不是很清楚呀。#include#define N 1100#define max 999999

#include int main(){ freopen("slyar.in", "r", stdin); freopen("slyar.out", "w", stdout);/**//* 中间按原样写代码，什么都不用修改 */ fclose(stdin); fclose(stdout);return 0;}

<br />#include<stdio.h> int perfect(int n) { int i,sum; sum=0; for(i=1;i<n/2+1;i++) if(n%i==0) sum+=i; if(sum==n) return 1; else return 0; } int main() { int i,n,count,a,b,t; scanf("%d",&n); while(n--) { count=0; s