题目描述
Given are a sequence of N positive integers A1,A2,…,AN, and a positive integer K.
Find the number of non-empty contiguous subsequences in A such that the remainder when dividing the sum of its elements by K is equal to the number of its elements. We consider two subsequences different if they are taken from different positions, even if they are equal sequences.
Constraints
·All values in input are integers.
·1≤N≤2×105
·1≤K≤109
·1≤Ai≤109
输入
Input is given from Standard Input in the following format:
N K A1 A2 ⋯ AN
输出
Print the number of subsequences that satisfy the condition.
样例输入
【样例1】 5 4 1 4 2 3 5 【样例2】 8 4 4 2 4 2 4 2 4 2 【样例3】 10 7 14 15 92 65 35 89 79 32 38 46
样例输出
【样例1】 4 【样例2】 7 【样例3】 8
提示
样例1解释
Four sequences satisfy the condition: (1), (4,2), (1,4,2), and (5).
样例2解释
(4,2) is counted four times, and (2,4) is counted three times.
#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector,fast-math") #pragma GCC optimize("Ofast") #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native") #pragma comment(linker, "/stack:200000000") #pragma GCC optimize (2) #pragma G++ optimize (2) #include <bits/stdc++.h> #include <algorithm> #include <map> #include <queue> #include <set> #include <stack> #include <string> #include <vector> using namespace std; #define wuyt main typedef long long ll; #define HEAP(...) priority_queue<__VA_ARGS__ > #define heap(...) priority_queue<__VA_ARGS__,vector<__VA_ARGS__ >,greater<__VA_ARGS__ > > template<class T> inline T min(T &x,const T &y){return x>y?y:x;} template<class T> inline T max(T &x,const T &y){return x<y?y:x;} //#define getchar()(p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++) //char buf[(1 << 21) + 1], *p1 = buf, *p2 = buf; ll read(){ll c = getchar(),Nig = 1,x = 0;while(!isdigit(c) && c!='-')c = getchar(); if(c == '-')Nig = -1,c = getchar(); while(isdigit(c))x = ((x<<1) + (x<<3)) + (c^'0'),c = getchar(); return Nig*x;} #define read read() const ll inf = 1e15; const int maxn = 2e5 + 7; const int mod = 1e9 + 7; #define start int wuyt() #define end return 0 int num[maxn],num2[maxn],n,k; map<int,int> mp; ll ans; start { n=read,k=read; for(int i=1;i<=n;++i) { num[i]=read; num2[i]=(num2[i-1]+num[i])%k; } for(int i=1;i<=n;++i) num2[i]=((num2[i]-i)%k+k)%k;//负数 for(int i=0;i<=n;++i) { if(i-k>=0) mp[num2[i-k]]--; ans+=mp[num2[i]]; mp[num2[i]]++; } cout<<ans<<endl; end; } /************************************************************** Language: C++ Result: 正确 Time:202 ms Memory:12964 kb ****************************************************************/
