对于每一个i找到最近的j满足最大值-最小值>K,对答案的贡献为j-i,用单调队列维护最值即可
#include<iostream> #include<cstdlib> #include<cstring> #include<cstdio> #include<algorithm> #include<queue> #include<cmath> #include<map> #define ll long long using namespace std; const int maxn=500010,inf=1e9; int n,m,l,r,L,R,K; int q[maxn],Q[maxn],a[maxn]; ll ans; void read(int &k) { int f=1;k=0;char c=getchar(); while(c<'0'||c>'9')c=='-'&&(f=-1),c=getchar(); while(c<='9'&&c>='0')k=k*10+c-'0',c=getchar(); k*=f; } int main() { read(n);read(K); for(int i=1;i<=n;i++)read(a[i]); l=1;r=0;L=1;R=0; for(int i=1,j=1;i<=n;i++) { while(j<=n) { while(l<=r&&a[q[r]]>=a[j])r--; while(L<=R&&a[Q[R]]<=a[j])R--; q[++r]=Q[++R]=j; if(a[Q[L]]-a[q[l]]>K)break; j++; } ans+=j-i; q[l]==i&&(l++);Q[L]==i&&(L++); } printf("%lld ",ans); return 0; }