[USACO16JAN]子共七Subsequences Summing to Sevensa[i]表示前缀和如果a[i]%7==t&&a[j]%7==t那么a[j]-a[i-1]一定是7的整数倍,这样就o(n)扫一遍,不断更新答案就可以了. #include<iostream> #include<cstdio> #include<queue> #include<algorithm> #include<cmath> #include…

P3131 [USACO16JAN]子共七Subsequences Summing to Sevens 题目描述 Farmer John's NN cows are standing in a row, as they have a tendency to do from time to time. Each cow is labeled with a distinct integer ID number so FJ can tell them apart. FJ would like to t…

P3131 [USACO16JAN]子共七Subsequences Summing to Sevens 题目描述 Farmer John's cows are standing in a row, as they have a tendency to do from time to time. Each cow is labeled with a distinct integer ID number so FJ can tell them apart. FJ would like to take…

Subsequences Summing to Sevens 题目描述 Farmer John's N cows are standing in a row, as they have a tendency to do from time to time. Each cow is labeled with a distinct integer ID number so FJ can tell them apart. FJ would like to take a photo of a conti…

题目大意:给个序列,求最长的连续子序列使其为7的倍数 又是一道令人欢喜的不用怎么用脑的水题.. 边读入,边计算前缀和 分别保存前缀和%7结果为1,2,3,4,5,6的第一次的位置 然后减一减就知道长度啦. #include<stdio.h> #include<string.h> #include<algorithm> using namespace std; ],ans; ]; int main(){ scanf("%d", &n); ans…

前缀和. 设f[i]为前缀和%7=i的第一个点.那么答案就是max(i-f[s[i]%7])了. #include<cstdio> #include<algorithm> #include<cstring> using namespace std; + ; int a[maxn],s[maxn]; ],n,ans; int main() { ;i<;i++) f[i]=-; scanf("%d",&n); ;i<=n;i++) {…

看到这一题第一印象就是暴力好打,$O(n^2)$,预计得分$70$分 这明显满足不了啊,我们要用到前缀和. $sum[i]$记录到i的前缀和,区间$[a,b]$的和就是$sum[b]-sum[a-1]$. 处理完以后怎么统计呢,$n^2$当然不行,我们要用到一个显然的定理. 如果 $a\equiv c(mod$ $k)$并且$b\equiv c(mod$ $k)$,那么$|a-b|\equiv 0(mod$ $k)$ 显然两个数的余数在相减的时候同时减去,从而只剩下$k$的倍数. 所以题目里我们…

https://www.luogu.org/problemnew/show/P3131 A表示前缀和数组 A[r] - A[l - 1] = 0 (mod 7) 得 A[r] = A[l - 1] (mod 7) #include <cstdio> #include <algorithm> using namespace std; ; #define yxy getchar() ], last[]; inline int read() { , f = ; char c = yxy;…

用sum[i]表示前缀和模7的值,若存在i≤j,满足sum[i]==sum[j],则区间(i,j]的和为7的倍数. O(N)扫出sum[0]~sum[6]第一次出现的位置first和最后一次出现的次数last,答案即为Max(last[i]-first[i]) #include<cstdio> #include<algorithm> using namespace std; ; ],last[]; void read(int &k){ k=; ; char c=getcha…

家里蹲大学数学杂志[官方网站]从由赣南师范大学张祖锦老师于2010年创刊;每年一卷, 自己有空则出版, 没空则搁置, 所以一卷有多期.本杂志至2016年12月31日共7卷493期, 6055页.既然做了, 就必须对自己和各位同学负责, 本杂志利用Latex精心排版, 整齐美观; 利用所学所知, 证明简单明了, 思路清晰;利用软件验算, 解答过程清楚, 结果准确. 从2017年起本刊除非应邀给出试题解答, 极少更新, 而逐步向``跟锦数学’’和``数学分析高等代数考研试题参考解答’’转换. 本杂志…