HDU OJ 4334 Trouble 2012 Multi-University Training Contest 4
题目：click here
题意：
给定5组数据，每组数据选择一个数，看是否能找到5个数的和为零。
分析：
千万不要~~T~~
普通线性查找：
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 6 using namespace std; 7 typedef long long ll; 8 const int INF = 0x3f3f3f3f; 9 const int M = 205; 10 11 int t, n; 12 ll a[5][M]; // 5组数据 13 ll pre[M*M]; // 第一二组数据两两和 14 ll nex[M*M]; // 第三四组数据两两和 15 ll lef[M]; // 最后一组数据 16 17 void solve() { 18 scanf("%d", &n ); 19 for( int i=0; i<5; i++ ) 20 for( int j=0; j<n; j++ ) 21 scanf("%I64d", &a[i][j] ); 22 int pos = 0; 23 for( int i=0; i<n; i++ ) 24 for( int j=0; j<n; j++ ) 25 pre[pos++] = a[0][i] + a[1][j]; 26 pos = 0; 27 for( int i=0; i<n; i++ ) 28 for( int j=0; j<n; j++ ) 29 nex[pos++] = a[2][i] + a[3][j]; 30 ll maxn = n*n; 31 for( int i=0; i<n; i++ ) 32 lef[i] = a[4][i]; 33 sort( pre, pre+maxn ); // 排序 34 sort( nex, nex+maxn ); 35 sort( lef, lef+n ); 36 for( int i=0; i<n; i++ ) { 37 int left = 0, right = maxn-1; 38 while( left<maxn && right >=0 ) { // 两个指针最多改变40000+40000次 39 ll sum = pre[left] + nex[right]; 40 if( sum == -lef[i] ) { 41 printf("Yes\n"); return ; 42 } 43 else if( sum > -lef[i] ) right--; 44 else left++; 45 } 46 } 47 printf("No\n"); 48 } 49 50 int main() { 51 scanf("%d", &t ); 52 while( t-- ) 53 solve(); 54 }
hash：(第一次碰到hash的题---不解释~~我也不懂)
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include <set> 6 7 using namespace std; 8 typedef __int64 ll; 9 const int INF = 0x3f3f3f3f; 10 const int MOD = 4e5+7; 11 const int M = 205; 12 13 int t, n; 14 ll a[5][M]; 15 ll g[MOD]; 16 ll sum; 17 bool f[MOD]; 18 19 int fhash( ll x ) { 20 int p = x%MOD; 21 if( p < 0 ) p += MOD; 22 while( f[p] && g[p] != x ) 23 p = (p+1)%MOD; 24 return p; 25 } 26 void solve() { 27 scanf("%d", &n ); 28 for( int i=0; i<5; i++ ) 29 for( int j=0; j<n; j++ ) 30 scanf("%I64d", &a[i][j] ); 31 memset( f, false, sizeof(f) ); 32 for( int i=0; i<n; i++ ) { 33 for( int j=0; j<n; j++ ) { 34 sum = a[0][i] + a[1][j]; 35 int p = fhash( sum ); 36 f[p] = true; 37 g[p] = sum; 38 } 39 } 40 for( int i=0; i<n; i++ ) { 41 for( int j=0; j<n; j++ ) { 42 for( int k=0; k<n; k++ ) { 43 sum = a[2][i] + a[3][j] + a[4][k]; 44 int p = fhash( -sum ); 45 if( f[p] ) { 46 printf("Yes\n"); 47 return ; 48 } 49 } 50 } 51 } 52 printf("No\n"); 53 } 54 int main() { 55 scanf("%d", &t ); 56 while( t-- ) 57 solve(); 58 }
