Starting from point (0,0) on a plane, we have written all non-negative integers 0, 1, 2,... as shown in the figure. For example, 1, 2, and 3 has been written at points (1,1), (2,0), and (3, 1) respectively and this pattern has continued. You are to w…

题意:找规律 思路:找规律 #include<iostream> #include<stdio.h> using namespace std; int main(){ int n,x,y; scanf("%d",&n); while(n--){ scanf("%d%d",&x,&y); ?printf():printf("%d\n",x+x); )x&?printf():printf(&qu…

Problem Description Starting from point (0,0) on a plane, we have written all non-negative integers 0, 1, 2,- as shown in the figure. For example, 1, 2, and 3 has been written at points (1,1), (2,0), and (3, 1) respectively and this pattern has conti…

http://acm.hdu.edu.cn/showproblem.php?pid=1391 Number Steps Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 3280    Accepted Submission(s): 2030 Problem Description Starting from point (0,0) on…

HDU 1711 Number Sequence(数列) Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) [Description] [题目描述] Given two sequences of numbers : a[1], a[2], ...... , a[N], and b[1], b[2], ...... , b[M] (1 <= M <= 10000, 1 <= N…

HDU 1005 Number Sequence(数列) Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) [Description] [题目描述] A number sequence is defined as follows: f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7. Given A, B, an…

Number Steps Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4482    Accepted Submission(s): 2732 Problem Description Starting from point (0,0) on a plane, we have written all non-negative integ…

HDU 1005 Number Sequence(数论) Problem Description: A number sequence is defined as follows:f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7. Given A, B, and n, you are to calculate the value of f(n).   Input The input consists of multipl…

HDU 4054 Number String 思路: 状态:dp[i][j]表示以j结尾i的排列 状态转移: 如果s[i - 1]是' I ',那么dp[i][j] = dp[i-1][j-1] + dp[i-1][j-2] + .. + dp[i-1][1] 如果s[i - 1]是‘D’,那么dp[i][j] = dp[i-1][j] + dp[i-1][j+1] + ... + dp[i-1][i] 用前缀和处理出sum[i][j]就不用dp[i][j]了 代码: #include<bits…

HDU 1711 Number Sequence (字符串匹配,KMP算法) Description Given two sequences of numbers : a1, a2, ...... , aN, and b1, b2, ...... , bM (1 <= M <= 10000, 1 <= N <= 1000000). Your task is to find a number K which make aK = b1, aK+1 = b2, ...... , aK+M…