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hdu-5014-Number Sequence-XiAn网络赛1008-水题
思路:既然是求两个数的异或运算之和,且由于数字不重复,那么肯定两个数异或的结果数字越大越好,即异或后从ai二进制的最高位后全是1。
具体思路看代码:
AC代码:
1 #include <iostream> 2 #include <cstdio> 3 #include <queue> 4 #include <vector> 5 #include <algorithm> 6 using namespace std; 7 int bit[17]; 8 int brr[100005]; 9 int n;10 void init()11 {12 bit[0] = 1;13 for(int i = 0; i <= 16; i++) {14 bit[i+1] = 2<<i;15 }16 }17 int arr[100005];18 void deal(int x)19 {20 for(int i = 0; i < n+10; i++) arr[i] = -1;21 arr[0] = 0;22 int k, le, xx;23 while(x != 0){24 if(x <= 0) return;25 int pos = upper_bound(bit, bit+18, x) - bit - 1;26 if(pos == 0) {27 arr[0] = 1; arr[1] = 0;28 return;29 }30 xx = x;31 if(pos == 1) k = 0x00000003 ^ x;32 else if(pos == 2) k = 0x00000007 ^ x;33 else if(pos == 3) k = 0x0000000f ^ x;34 else if(pos == 4) k = 0x0000001f ^ x;35 else if(pos == 5) k = 0x0000003f ^ x;36 else if(pos == 6) k = 0x0000007f ^ x;37 else if(pos == 7) k = 0x000000ff ^ x;38 else if(pos == 8) k = 0x000001ff ^ x;39 else if(pos == 9) k = 0x000003ff ^ x;40 else if(pos == 10) k = 0x000007ff ^ x;41 else if(pos == 11) k = 0x00000fff ^ x;42 else if(pos == 12) k = 0x00001fff ^ x;43 else if(pos == 13) k = 0x00003fff ^ x;44 else if(pos == 14) k = 0x00007fff ^ x;45 else if(pos == 15) k = 0x0000ffff ^ x;46 else if(pos == 16) k = 0x0001ffff ^ x;47 le = k;48 x = k - 1;49 for(int i = xx; i >= k; i--) arr[i] = le++;50 }51 }52 int main()53 {54 init();55 while(scanf("%d", &n) != EOF) {56 deal(n);57 long long ans = 0;58 for(int i = 0; i <= n; i++) {59 int a; scanf("%d", &a);60 brr[i] = arr[a];61 ans += a ^ arr[a];62 }63 printf("%I64d\n%d", ans, brr[0]);64 for(int i = 1; i <= n; i++) {65 printf(" %d", brr[i]);66 }67 printf("\n");68 }69 return 0;70 }
总结:QAQ比赛的时候看错题意了。。。。忽略了数字不重复的条件,导致想了好久。。。。
hdu-5014-Number Sequence-XiAn网络赛1008-水题
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