洛谷P3131 [USACO16JAN] Subsequences Summing to Sevens S(前缀和+模运算性质) 做题历程 拿到手的时候就是考虑前缀和,毕竟要求区间和,如果暴力做就是N3,那么就开始做,写了个O(n2)复杂度代码,交上去80分...迷了,心想这题难道dp啊,懒得想的我直接看题解,发现大佬用了一个很简单的模运算性质就过去了,即(a...
给你n个数,分别是a[1],a[2],...,a[n]。求一个最长的区间[x,y],使得区间中的数(a[x],a[x+1],a[x+2],...,a[y-1],a[y])的和能被7整除。输出区间长度。若没有符合要求的区间,输出0。 输入格式 The first line of input containsNN (1 \leq N \leq 50,0001≤N≤50,000). The n...
洛谷P3131 [USACO16JAN]子共七Subsequences Summing to Sevens,P3131[USACO16JAN]子共七SubsequencesSummingtoSevensP3131[USACO16JAN]子共七SubsequencesSummingtoSevensP3131[USACO16JAN]子共七SubsequencesSummingtoS
sum[i]%=7; }for(inti=1;i<=n;i++) {if(!f[sum[i]])f[sum[i]]=i;elseans=max(ans,i-f[sum[i]]); } printf("%d",ans); }
洛谷P3131 Subsequences Summing to Sevens S 题目大意#给你nn个数,分别是a[1],a[2],...,a[n]a[1],a[2],...,a[n] 求一个最长的区间 [x,y][x,y],使得区间中的数a[x],a[x+1],a[x+2],...,a[y−1],a[y]a[x],a[x+1],a[x+2],...,a[y−1],a[y]的和能被7整除...
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 a photo of a contiguou...
洛谷P3131 [USACO16JAN]Subsequences Summing to Sevens S 题目描述# Copy 给你n个数,分别是a[1],a[2],...,a[n]。求一个最长的区间[x,y],使得区间中的数(a[x],a[x+1],a[x+2],...,a[y-1],a[y])的和能被7整除。 输出区间长度。若没有符合要求的区间,输出0。
洛谷P3131 [USACO16JAN]子共七Subsequences Summing to Sevens 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 ...