心在路上

解题思路：本题和最右边数类似，找规律，四个一循环 /*A hard puzzle*/ #include using namespace std; int main(int argc, char *argv[]) { int m,n; int a,b,c,d; while(cin>>m>>n) { a=m%10; b=(a*a)%

Problem Description Ignatius is poor at math,he falls across apuzzle problem,so he has no choice but to appeal to Eddy. this problemdescribes that:f(x)=5*x^13+13*x^5+k*a*x,input a nonegative integerk

#include int main(int argc, char *argv[]) { int i; double n,sum; while(scanf("%lf",&n)!=EOF) { if(n==0) break; for(i=2,sum=0; sum<n; i++) sum+=1.0/i;

Encoding Time Limit: 2000/1000 MS (Java/Others)    MemoryLimit: 65536/32768 K (Java/Others) Total Submission(s): 16720    Accepted Submission(s): 7132 Problem Description Givena string conta

题目大意：跳格子，约束条件是，下一步格子的数比上一步格子的数要大。 解题思路：典型的最长有序子序列，并记录各个序列的sum，输出最大的sum；   /*hdu1087*/ #include #define nMax 1024 typedef struct data { long n; long sum; }; struct data a[nMax]; int main

Letthe Balloon Rise Time Limit: 2000/1000 MS (Java/Others)    MemoryLimit: 65536/32768 K (Java/Others) Total Submission(s): 44096    Accepted Submission(s): 15575 Problem Description Contest

FibonacciAgain Time Limit: 2000/1000 MS (Java/Others)    MemoryLimit: 65536/32768 K (Java/Others) Total Submission(s): 21563    Accepted Submission(s): 10360 Problem Description There areano

Ignatiusand the Princess II Time Limit: 2000/1000 MS (Java/Others)    MemoryLimit: 65536/32768 K (Java/Others) Total Submission(s): 2675    Accepted Submission(s): 1622 Problem Description N

题目大意：机器人走迷宫，给定一张地图，机器人根据地图的方向走，能走出地图的则输出走过的步数，循环走的，根据题目输出循环走前的步数以及循环走的步数 解题思路：dfs模拟机器人走； /*Robot Motion*/ #include #include using namespace std; int row,col,start; char gmap[11][11]; int visit[11

题目大意：放置炮台，两个炮台在同行或者同列，则不能直接摆置，中间要有墙才能放置； 解题思路：用dfs找出每种摆置炮台的方法，记录最多炮塔数，本题的图规格比较小，所以dfs耗时不高。 /*HDU 1045 Fire Net*/ #include #include #include using namespace std; char gmap[4][4]; int maxNum=0; in

解题思路：看到N的范围，可以断定本题是找规律的，可以从最小的2的1次方到4次方，2,4,8,16，其最右边数分别是2,4,5,6，可以看出，任意M>=2的数的N次方，会在4次以内有循环。   #include int main() { int n,i,j,k,l; long m; scanf("%d",&n); while(n--) {

字符串的倒置，注意字符串接受和个别细节就ok了，接受cas的时候，下面接受字符用getchar（）接受换行符，否则换行符会当输入数据接收的。   #include #include #include #define nMax 1024 using namespace std; string s1; char s[nMax]; char c; int main(int argc, cha

Number Sequence Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 6786    Accepted Submission(s): 3084 Problem Description Given two