原題:http://acm.hdu.edu.cn/showproblem.php?pid=2141
題目大意:輸入A,B,C三個整型數組(數組長度1<=len<=500),然後訪問S(1<=S<=1000)次,判斷A,B,C三個數組中是否存在這樣三個數使得A[i]+B[j]+C[k] = X;是就輸出YES,否則輸出NO;
分析:如果直接將三個數組合並最壞的情況是500^3的數組長度,而且三個整型數相加,減很有可能就超過了int的範圍了。所以還要另想辦法,如果把等式變形A[i]+B[j] = X –C[k];(兩個int 數相加減也有可能超過範圍,這就要看測試數據了),這樣變形後,就可以把A,B數組合併成ab[500*500],然後在ab[]中找X-C[k],這裏可以選擇用二分查找,查找過程還可以進行一些優化……
代碼:
#include <stdio.h>
#include <algorithm>
#include <string.h>
using namespace std;
int num_a[510],num_b[510],num_c[510],a,b,c,len;
int ab[250000];
void addab(int num_a[],int num_b[])
{
len = 1;
memset(temp,0,sizeof(temp));
for (int i = 0 ; i < a; i++) // 合併num_a,num_b
for (int j = 0; j < b; j++)
ab[len++] = num_a[i] + num_b[j];
sort(ab+1,ab+len);
}
bool find(int s)
{
int low,heigh,mid,flag,n;
if (s - num_c[c-1] > ab[len-1] ) return false; //如果要找的s比最大值還大
if (s - num_c[0] < ab[1] ) return false; //或s比最小值還小
for (int k = 0; k < c; k++)
{
n = s-num_c[k];
flag = 0;
if (n >= ab[1] && n <= ab[len-1])
{
low = 1,heigh = len-1, flag = 1;
while (low <= heigh) //在ab中二分查找n;
{
mid = (low + heigh)/2;
if (ab[mid] < n)
low = mid+1;
else if (ab[mid] > n)
heigh = mid-1;
else if (ab[mid] == n)
return true;
}
}
}
return false;
}
int main()
{
int i,case_n,j,k,n,s,Max_a = -1,Max_b = -1;
case_n = 0;
while (scanf("%d%d%d",&a,&b,&c) != EOF)
{
case_n++;
for (i = 0; i < a; i++) scanf("%d",&num_a[i]);
for (j = 0; j < b; j++) scanf("%d",&num_b[j]);
for (k = 0; k < c; k++) scanf("%d",&num_c[k]);
sort(num_c+0,num_c+c);
addab(num_a,num_b); //將num_a,num_b合併存入ab中
scanf("%d",&n);
printf("Case %d:\n",case_n);
while (n > 0)
{
scanf("%d",&s);
if (find(s))
printf("YES\n");
else
printf("NO\n");
n--;
}
}
return 0;
}