pusidun的专栏

待定系数，一般这种题平面设f(n)=

先存下#includeint a[1001],dp[1001];int main(){ int n,i,j,max,maxn; while(scanf("%d",&n)!=EOF&&n!=0) { for(i=1;i<=n;i++){ scanf("%d",&a[i]); dp[i]=a[i]; } maxn=-1; for(i=2;i<=n;i++

#includeint main(){ int n; char ch; while(scanf("%d",&n)!=EOF&&n!=0) { getchar(); int count=0; while(n--) { while((ch=getchar())!='\n') { if(ch>='0'&&ch<='9') count++; }

#includeint main(){ int n,a[100]; while(scanf("%d",&n)!=EOF) { if(n==0)break; for(int i=0;i<n;i++) scanf("%d",&a[i]); int min=a[0],j=0; for(int i=0;i<n;i++) if(min>a[i]) {

水仙花数Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 92329    Accepted Submission(s): 27357Problem Description春天是鲜花的季节，水仙花就是其中最迷人的代表，数学上有个水仙花

#includeint main(){ int n; while(scanf("%d",&n)==1) { int sum=1; for(int i=n;i>1;i--) sum=2*(sum+1); printf("%d\n",sum); } return 0;}

#includeint is_prime(int n){ int i; if(n<2) return 0; for(i=2;i*i<n;i++) { if(n%i==0) return 0; } return 1;}int main(){ int x,y; while(scanf("%d%d",&x,&y)==2&&(x!=0||y!=0)) { int cou

#includevoid fun(int n){ int j,count=1,q; double sum=0; for(j=1;j<=n;j++) { if(count%2==1)q=1; else q=-1; sum+=1.0*q/j; count++; } printf("%.2lf\n",sum);}int main(){ int m,n,i; sc

数学问题，用对数求解log(a*b)

Problem ATime Limit : 1000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)Total Submission(s) : 38   Accepted Submission(s) : 16Problem Description在讲述DP算法的时候，一个经典的例子就是数塔

#include#includechar trans(char c){ if(c>=65&&c<=69) return c+21; else if(c>69&&c<=91) return c-5; else return c;}using namespace std;int main(){ string beg,end,str; while(1) { ge

递推题目设第n年有羊f(n)，其中有f(n-1)个前面留下来的羊，

这题目背景略长啊，你要是先看到我这个

用函数递归又超时了= =直接用的数组存

#include#includeint main(){ char a[100]; while(gets(a))//use gets(); { int n=strlen(a); a[0]=a[0]-32; for(int i=1;i<n-1;i++) { if(a[i]==' ') a[i+1]-=32; } printf("%s\n",a); }