题意：求sum=n/1+n/2+n/3+...+n/n。(n<2^31)
分析：在一定的区间内n/i的值是一定的，因此要跳过这段区间来加速求解。
#pragma comment(linker,"/STACK:1024000000,1024000000") #include <cstdio> #include <cstring> #include <string> #include <cmath> #include <limits.h> #include <iostream> #include <algorithm> #include <queue> #include <cstdlib> #include <stack> #include <vector> #include <set> #include <map> #define LL long long #define mod 100000000 #define inf 0x3f3f3f3f #define eps 1e-6 #define N 10000000 #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define PII pair<int,int> using namespace std; inline LL read() { char ch=getchar();LL x=0,f=1; while(ch>'9'||ch<'0'){if(ch=='-')f=-1;ch=getchar();} while(ch<='9'&&ch>='0'){x=x*10+ch-'0';ch=getchar();} return x*f; } int main() { int n; int T,cas=1; T=read(); while(T--) { n=read(); LL ans=0; for(LL i=1,last=0;i<=n;i=last+1) { last=n/(n/i); ans+=(last-i+1)*(n/i); } printf("Case %d: %lld ",cas++,ans); } }