hdu 5683 zxa and xor 暴力 zxa and xor
Time Limit: 16000/8000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Problem Description
zxa had a great interest in exclusive disjunction(i.e. XOR) recently, therefore he took out a non-negative integer sequence ).
Input
The first line contains an positive integer 5
Output
For each test case, output in i-th called function.
Sample Input
1
3 3
1 2 3
1 4
2 5
3 6
Sample Output
4
6
8
Hint
After the first called function, this sequence is ${4,2,3}$, and $(4+2)otimes(4+3)otimes(2+3)=4$.
After the second called function, this sequence is ${4,5,3}$ and $(4+5)otimes(4+3)otimes(5+3)=6$.
After the third called function, this sequence is ${4,5,6}$ and $(4+5)otimes(4+6)otimes(5+6)=8$.思路：通过位运算一些简单性质，暴力；（居然不会超时。。。）
#include<bits/stdc++.h> using namespace std; #define ll long long #define mod 1000000009 #define inf 999999999 #define esp 0.00000000001 //#pragma comment(linker, "/STACK:102400000,102400000") int scan() { int res = 0 , ch ; while( !( ( ch = getchar() ) >= '0' && ch <= '9' ) ) { if( ch == EOF ) return 1 << 30 ; } res = ch - '0' ; while( ( ch = getchar() ) >= '0' && ch <= '9' ) res = res * 10 + ( ch - '0' ) ; return res ; } int a[100010]; int main() { int x,y,z,i,t; scanf("%d",&x); while(x--) { scanf("%d%d",&y,&z); for(i=1;i<=y;i++) scanf("%d",&a[i]); int sum=0; for(i=1;i<=y;i++) for(t=i+1;t<=y;t++) sum^=(a[i]+a[t]); while(z--) { int pos,change; scanf("%d%d",&pos,&change); for(i=1;i<=y;i++) { if(i!=pos) sum^=(a[i]+change)^(a[i]+a[pos]); } printf("%d ",sum); a[pos]=change; } } return 0; }