Light OJ 1278 Sum of consecutive integers (number of odd factors)

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Test instructions: Divides the given n into successive numbers and has at least two numbers to see how many options are available.

Analysis:

A + (A + 1) + (A + 2) + ... +(A + k-1) = N;

===> (2*a + k-1) * k = 2*n;

===> (2*a-1) *k = 2*n-k*k;

===> 2*a-1 = 2*n/k-K;

The left side of the equation is odd, then the right side must be odd, then K must be an odd number of n

So the n factor decomposition is possible.

The code is as follows:

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include < cmath> #include <set> #include <map> #include <queue> #define PB push_back#define MP make_pair# Define REP (I,n) for (int i=0;i< (n), ++i) #define for (I,L,H) for (int i= (L), i<= (h), ++i) #define DWN (i,h,l) for (int i= (h ); i>= (l);-i) #define IFOR (I,H,L,V) for (int i= (h); i<= (L); i+= (v)) #define CLR (Vis) memset (vis,0,sizeof (VIS)) # Define MST (Vis,pos) memset (vis,pos,sizeof (VIS)) #define MAX3 (A,B,C) max (A,max (b,c)) #define MAX4 (A,B,C,D) max (max (b) , Max (c,d)) #define MIN3 (a,b,c) min (a,min (b,c)) #define MIN4 (a,b,c,d) min (min (A, b), Min (c,d)) #define PI ACOs (-1.0) # Define INF 1000000000#define linf 1000000000000000000ll#define eps 1e-8#define LL long longusing namespace Std;const int m    AXN = 1e7+10;int p[maxn/10],num;bool vis[maxn];void init () {num=0;    CLR (VIS);            for (i,2,maxn-1) {if (!vis[i]) {p[num++]=i;        IFOR (J,i+i,maxn-1,i) vis[j]=1; }   }}int Main () {init ();    int T,cas=1;    scanf ("%d", &t);        while (t--) {LL n;        scanf ("%lld", &n);        LL ans = 1;                for (int i=0;i<num&&p[i]<=n;i++) {if (n%p[i]==0) {int cnt = 0;                while (n%p[i]==0) n=n/p[i],cnt++;            if (p[i]&1) ans=ans* (LL) (cnt+1);        }} if (n>1&&n&1) ans=ans*2ll;        ans--;    printf ("Case%d:%lld\n", Cas++,ans); } return 0;}

Light OJ 1278 Sum of consecutive integers (number of odd factors)