题解:直接用公式算，用容斥来减掉重复计算的部分
但是我犯了一个非常sb的错误，直接把abcd除k了，这样算a-1的时候就错了，然后举的例子刚好还没问题= = ，结果wa了好几发
//#pragma comment(linker, "/stack:200000000") //#pragma GCC optimize("Ofast,no-stack-protector") //#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native") //#pragma GCC optimize("unroll-loops") #include<bits/stdc++.h> #define fi first #define se second #define mp make_pair #define pb push_back #define pi acos(-1.0) #define ll long long #define mod 1000000007 #define C 0.5772156649 #define ls l,m,rt<<1 #define rs m+1,r,rt<<1|1 #define pil pair<int,ll> #define pii pair<int,int> #define ull unsigned long long #define base 1000000000000000000 #define fio ios::sync_with_stdio(false);cin.tie(0) using namespace std; const double g=10.0,eps=1e-12; const int N=50000+10,maxn=400000+10,inf=0x3f3f3f3f; int mu[N],prime[N],sum[N]; bool mark[N]; int cnt; void init() { mu[1]=1; cnt=0; for(int i=2;i<N;i++) { if(!mark[i])prime[++cnt]=i,mu[i]=-1; for(int j=1;j<=cnt;j++) { int t=i*prime[j]; if(t>N)break; mark[t]=1; if(i%prime[j]==0){mu[t]=0;break;} else mu[t]=-mu[i]; } } for(int i=1;i<N;i++)sum[i]=sum[i-1]+mu[i]; } ll cal(int n,int m) { ll ans=0; for(int i=1,last;i<=min(n,m);i=last+1) { last=min(n/(n/i),m/(m/i)); ans+=(ll)(sum[last]-sum[i-1])*(n/i)*(m/i); } return ans; } int main() { init(); int t; scanf("%d",&t); while(t--) { ll a,b,c,d,k; scanf("%lld%lld%lld%lld%lld",&a,&b,&c,&d,&k); printf("%lld\n",cal(b/k,d/k)-cal((a-1)/k,d/k)-cal((c-1)/k,b/k)+cal((a-1)/k,(c-1)/k)); } return 0; } /******************** ********************/