多校 hdu
阿新 • • 發佈:2018-01-15
and fine php chm cst to do seve 3.6 ext
Solve this interesting problemTime Limit: 2000/1000 MS (Java/Others)????Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1571????Accepted Submission(s): 454 Problem Description Have you learned something about segment tree? If not, don’t worry, I will explain it for you. - For each node u in Segment Tree, u has two values:?Lu?and?Ru. - If?Lu=Ru, u is a leaf node.? - If?Lu≠Ru, u has two children x and y,with?Lx=Lu,Rx=?Lu+Ru2?,Ly=?Lu+Ru2?+1,Ry=Ru. Here is an example of segment tree to do range query of sum. Given two integers L and R, Your task is to find the minimum non-negative n satisfy that: A Segment Tree with root node‘s value?Lroot=0?and?Rroot=n?contains a node u with?Lu=L?and?Ru=R. ? Input The input consists of several test cases.? Each test case contains two integers L and R, as described above. LR?L+1≤2015 ? Output For each test, output one line contains one integer. If there is no such n, just output -1. ? Sample Input 6 7 10 13 10 11 ? Sample Output 7 -1 12 ? Source 2015 Multi-University Training Contest 3 ? Recommend wange2014???|???We have carefully selected several similar problems for you:?? pid=5326" style="color:rgb(26,92,200);text-decoration:none;">5326? pid=5325" style="color:rgb(26,92,200);text-decoration:none;">5325?5324? pid=5322" style="color:rgb(26,92,200);text-decoration:none;">5322?5321? ? |
#include<cstdio> #include<cmath> #include<stdlib.h> #include<map> #include<set> #include<time.h> #include<vector> #include<queue> #include<string> #include<string.h> #include<iostream> #include<algorithm> using namespace std; #define eps 1e-8 #define INF 0x3f3f3f3f #define LL long long #define max(a,b) ((a)>(b)?(a):(b)) #define min(a,b) ((a)<(b)?(a):(b)) LL L, R; LL flag; void dfs(LL l, LL r) { if((flag && r >= flag)) return ; if(l == 0) { flag == 0 ? flag = r : flag = min(flag, r); return ; } long long t = r - l + 1; if(t <= l) { dfs(l - t - 1, r); dfs(l - t, r); dfs(l, r + t - 1); dfs(l, r + t); } } int main() { while(~scanf("%I64d%I64d", &L, &R)) { flag = 0; if(R == 0) printf("0\n"); else { dfs(L, R); if(flag) printf("%I64d\n", flag); else printf("-1\n"); } } return 0; }
多校 hdu