zck921031的专栏

Qwerty the Ranger took up a government job and arrived on planet Mars. He should stay in the secret lab and conduct some experiments on bacteria that have funny and abnormal properties. The job isn't

//花束摆放问题，类似于背包，方程为f[i][j]=max{f[i][j-1],f[i-1][j-1]+a[i][j]} i=1..F, j=i+1..V//初始f[i][i]预先算出来, f[0][]={0}, 结果为f[F][V]//F为花朵数，V为花瓶数#include #include const int S=105;int a[S][S],f[S][S];in

递推式a[i+1]=a[i]*(a[i]+1)  输出的是a[i]+1要用高精度，最后答案在5.2w位左右。#include #include using namespace std;#define DIGIT 4#define DEPTH 10000#define MAX 15000typedef int bignum_t[MAX+1];void

map水过2400ms+懒得改成hash了 DescriptionThe Recaman's sequence is defined by a0 = 0 ; for m > 0, am = am−1 − m if the rsulting am is positive and not already in the sequence, otherwise am = am−1 + m.

学会打表。。。其实程序计算也挺快的，left=n-i; j=(m-1)%left+1;这两句加上去1s内就把表打好了。经典约瑟夫问题，直接用数组做的。 #include #include using namespace std;int h[40];int check(int n,int m,int k){ memset(h,0,sizeof(h));

超水，看懂题目很重要111%3=0111111%7=0...#include using namespace std;int main(){    int n,i,y;    while (cin>>n){        y=1;i=1;        while (y!=0){            i++;            y=(y*10+

有一艘船只能载两个人，总共有n个人要过河，每个人过河需要时间t[i], 两人共同过河时间为 max(t[i],t[j])，求过河时间最短。每次载最慢的两个人过河都有两种方法。 可以是t[0] t[1] 一起，也可以是t[0]一个人来回载。 #include #include using namespace std;#define MAX(x,y) (x<y?y:x)#d

求个最大值最小值，right on的时候判断一下是否合法就可以了。 #include #include using namespace std;char buff[30];int main(){    int n,max,min,x;    while (1){        min=-1;        max=100;        while (1

http://poj.org/problem?id=2707 #include #define MIN(x,y) (x<y?x:y)#define MAX(x,y) (x<y?y:x)int main(){ int a,b,c,d; double t1,t2,t; while (scanf("%d %d %d %d",&a,&b,&c,&d),a!

http://poj.org/problem?id=2739#include int plist[10000],pcount=0;int prime(int n){ int i; if ((n!=2&&!(n%2))||(n!=3&&!(n%3))||(n!=5&&!(n%5))||(n!=7&&!(n%7))) return 0; for (i=0;plis

就是将9进制转换为10进制，很简单http://poj.org/problem?id=2719#include int ch[10];int main(){ int x,i,sum,k,tmp; for (i=0;i<=3;i++) ch[i]=i; for (i=4;i<=9;i++) ch[i]=i-1; while (scanf("

Election TimeTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 5531 Accepted: 2998DescriptionThe cows are having their first election after overthrowing

看完题目后，得到公式a[i]=2*a[i-1]-a[i-2]+2*c[i-1];  YY发现a[n+1]随a[1]增大而增大，那么就对a[1]二分吧，精度1e-4,1A。Simple calculationsTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 5989 Accepte

只是简单的转换了一下进制，很容易MultiplyTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 4761 Accepted: 2551Description6*9 = 42" is not true for base 10, but is true for

用dfs算出每个连通块的大小即可DescriptionFarmer John purchased satellite photos of W x H pixels of his farm (1 <= W <= 80, 1 <= H <= 1000) and wishes to determine the largest 'contiguous' (connected) pasture. P

用最朴素的方法打出前三十组数据，发现是fibonacci数列，直接水过。此外还可以用排列组合的方法。n=3时，由插空法得C(4,0)+C(3,1)+C(2,2); n=k时则是sigma C(k-i+1,i)   (i=0..(k+1)/2) Problem Given a positive integer n, determine the number of different