Cheky_的专栏

#include int p(int a, int b) { if (b > a) { return 0; } else if (b == 0) { return 1; } else { int i = a - b + 1; int r = a; while (i < a) { r *= i; ++i; } return r; }}in

/*** * 以资源为列，用户为行，统计资源列有多少个不同值，去掉空值，即为答案。 */import java.util.*;import java.io.*;public class Solution {public static void main(String[] argv) throws IOException {BufferedReader in = new BufferedReader

f（n）表示问题的解 第一步：将n0 = n分成三部分a1 ，b1= 1， n1；其中a1 = n0 - n1 - 1，a1 〉=  n1；n1 是之后要分解的 ，可行解为 f（n1）我们得到  a1，b1， f（n1）前n1天，用n1（会拆开）支付，解为f（n1）第n1+1天用 b1支付第n1+2天用a1（完整的）支付，必须找零a1 - 1，店主有n1+1个，所以必须有

AC之后看大网上别人的基本是dfs、暴力搜索因为我的算法不是这样，贴出来看看sgu结果：  Accepted  31 ms   835 kb 英文很烂，注释看不懂可email /** * @file * @brief my c solution for problem 125 on SGU OJ * @author ck * @date 2013-06-19 *

#include #include const double PI = 3.1415926535897932384626;int n, n1, n2;double n1x, n1y, n2x, n2y;int main() { scanf("%d%d%d", &n, &n1, &n2); scanf("%lf%lf%lf%lf", &n1x, &n1y, &n2x, &n2

/* if n = a*b + r and gcd(n, r) = 1 then ans = max{a, b} (a, b <= n/2) if n = 2p + 1 then ans = p; else n = 2p let p = 2q+r if r=0 then n = 4q =2(2q - 1) + 2, gcd(n, 2q-1) = gcd(

/*notice: the graph may contain loops between two point,there may be many ways that have different color */#include #include using namespace std;const int INF = 0x7ffffffe;int n, m;int

/* 是某某的1/x 居然被翻译成： which is x times less than ... */#include int main() { int x; scanf("%d", &x); printf("%d\n", (x<<1) + 1); return 0;}

//dfs// k = 0， ans = 1；// k = 1， ans = n*n ；// k > n, ans = 0;//还可以进一步优化 // if (n - ps < k - count) rollback; 当剩余列数小于剩余棋子数时，回滚// 对称性优化, 第一个棋子放在某一列m行，和放在该列n-m+1// 直接用数组保存所有结果 ---

/* greedy method sort the leaflets by Li by descending order let sumTk = Ti + T2 + ..... Tk answer = max ( sumTk + Lk), k = 1...nproofproof by contradiction : 假设最优序列为 a = { 1, 2, 3 ..

/*状态压缩dp每一行的可行状态进行压缩 ， n=3时 s = {000， 001， 010， 100， 101}d[row][si][p] 表示 第row行状态为si，前row-1行已放置且不冲突，总共放了p个棋子的方案数初始状态 d[0][0][0] = 1， others = 0；递推公式 d[row][s[i]][p + c[i]] = sigma（d[row-1][s[j]

#include #define NONE 'n'/*升序数列，删除前面的高位降序数列，删除后面的低位先处理升序*/char n[1009];int k;// find the next digit after digit a, return -1 when can not find.int next(int a) {

题目中： Each shot always reduces the hit points of the target tank by 8, and if the number of hit points of the targetbefore the shot was greater than zero, the shooting tank receives 3 score points.

#include int main() { int n, m, i ; char s[109]; scanf("%d%d", &n, &m); n %= m; if (n==0) { n = m; } for (i=1; i<=m; ++i) { scanf("%s", s); if (i

#include #include void putUpper(char c) { if (c >= 'a') { //c is lowercase putchar(c - 32); } else { putchar(c); }}void putLower(char c) { if (c <= 'Z') {

#include int main() { int p, s; int t[] = {1, 300, 900, 1800, 10800 }; scanf("%d%d", &s, &p); p -= s; s = 0; while (t[s] <= p) ++s; printf("%d\n", s); return 0

做sgu oj的一些ac代码sgu地址 ： http://acm.sgu.ru/我的帐号   052435

//暴力求解复杂度为O（n(n-1)/2）//假设所有输入中最大值为maxp,枚举i=maxp to 1（从大到小枚举）,//如果i是某两个数的约数，则i是所有数的最大最大公约数。//复杂度 ？？#include #include #define maxn 1000005 int num[maxn]; int main() { int n,i,j,k;