Problem E. Arithmetic Sequence Input file: standard input Output file: standard output Time limit: 1 seconds Memory limit: 64 mebibytes A sequence b1, b2, . . . , bn are called (d1, d2)-arithmetic sequence if and only if there exist i (1 ≤ i ≤ n) such that for every j (1 ≤ j < i),bj+1 = bj + d1 and for every j(i ≤ j < n), bj+1 = bj + d2. Teacher Mai has a sequence a1, a2, . . . , an. He wants to know how many intervals [l, r] (1 ≤ l ≤ r ≤ n) there are that al , al+1, . . . , ar are (d1, d2)-arithmetic sequence. Input First line of the input contains one integer T (1 ≤ T ≤ 15) — number of test cases. For each test case, the first line contains three numbers n,d1,d2(1 ≤ n ≤ 105 ,|d1|, |d2| ≤ 1000), the next line contains n integers a1, a2, . . . , an (|ai| ≤ 109 ).
题目分析：
只和差分有关，存一个差分数组，找出形如 d1,d1...d1,d2,d2...d2,的统计方案数
#include<bits/stdc++.h> using namespace std; typedef long long ll; const int MAXN = 1e5+100; ll a[MAXN]; ll b[MAXN]; void Run() { int n; ll d1,d2; scanf("%d%lld%lld",&n,&d1,&d2); for(int i=0;i<n;i++) scanf("%lld",a+i); for(int i=0;i<n-1;i++) b[i]=a[i+1]-a[i]; int L,M,R; ll cnt1=0,cnt2=0; ll ans=n; n=n-2; //for(int i=0;i<=n;i++) cout<<b[i]<<endl; int i=0; while(i<=n) { while(i<=n&&(b[i]!=d1&&(b[i]!=d2))) i++; cnt1=0; while(i<=n&&b[i]==d1) cnt1++,i++; cnt2=0; while(i<=n&&b[i]==d2) cnt2++,i++; ans+=(cnt1*cnt2+(cnt1*cnt1+cnt1)/2+(cnt2*cnt2+cnt2)/2); } printf("%lld",ans); } int main() { int T; scanf("%d",&T); while(T--) { Run(); printf(" "); } return 0; }