rofn123

#include int arr[500000] = {};int tree[2000000] = {};int lazy[2000000] = {};int init(int s, int e, int node) { if(s==e) return tree[node] = arr[s]; int m=(s+e)/2; return tree[node] = init(s,m,node*2)^init(m+1,e,node*2+1);}void lazyUpdate(int s, int e, int node) { if(lazy[node]) { if(((e-s)&1)==0) tree[node] ^= lazy[node]; if(s!=e) { lazy[node*2] ^= lazy[node]; lazy[node*2+1] ^= lazy[node..

#include int arr[500000] = {};int tree[2000000] = {};int lazy[2000000] = {};int init(int s, int e, int node) { if(s==e) return tree[node] = arr[s]; int m=(s+e)/2; return tree[node] = init(s,m,node*2)^init(m+1,e,node*2+1);}void lazyUpdate(int s, int e, int node) { if(lazy[node]) { if(((e-s)&1)==0) tree[node] ^= lazy[node]; if(s!=e) { lazy[node*2] ^= lazy[node]; lazy[node*2+1] ^= lazy[node..

#include int f(int *a) { int b = *a%4; if(b==0) b=*a; else if(b==1) b=1; else if(b==2) b=*a+1; else if(b==3) b=0; return b; } int main(void) { int n; scanf("%d",&n); while(n--) { int a, b; scanf("%d %d", &a, &b); --a; printf("%d\n", f(&b)^f(&a)); } } XOR의 성질을 이용하는 문제입니다. 1. 1부터 연속적으로 XOR을 하면 0001 0001 ... 1 0002 0003 ... 3 = 2 + 1 0003 0000 ... 0 0004 0004 ... 4 0005 0001 ... 1 0006 0007 ... 7 =..

#include #include int main(void) { int n, m, ans = 0; scanf("%d %d", &n, &m); int *a = (int*) malloc(4*(n+1)); //동적할당 a[0] = 0; for(int i=1; i