hdu 2141 Can you find it(二分)
2013-12-12 11:38
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链接:http://acm.hdu.edu.cn/showproblem.php?pid=2141
题意:给出三个数列a,b,c和一组数x,求对于每个x是否存在 ai + bj + ck = x;其中三个数列最多各有500个数,x最多有1000个。
思路: 将数列a,b两两相加,组成一个数列(500*500)。
然后枚举数列c,利用二分从合并数列中查找满足条件的数。
注意事项:二分循环的条件为while(left<right) ,所以若跳出循环后left==right,要加一组判断。if(left==right && c[i]+d[left]==x) flag = 1;
1 #include <cstdio>
2 #include <cstring>
3 #include <algorithm>
4 using namespace std;
5 #define N 505
6 #define S 1005
7
8 int a
,b
,c
,d[N*N];
9
int main()
{
int l, n, m, s, x, Case = 0;
while(scanf("%d%d%d",&l,&n,&m)!=EOF)
{
printf("Case %d:\n",++Case);
for(int i=1; i<=l; i++)
scanf("%d",&a[i]);
for(int i=1; i<=n; i++)
scanf("%d",&b[i]);
for(int i=1; i<=m; i++)
scanf("%d",&c[i]);
int num = 0;
for(int i=1; i<=l; i++)
{
for(int j=1; j<=n; j++)
d[++num] = a[i] + b[j];
}
sort(d+1,d+num+1);
scanf("%d",&s);
while(s--)
{
scanf("%d",&x);
int flag = 0;
for(int i=1; i<=m; i++)
{
int left = 1, right = num, mid;
while(left<right)
{
mid = (left + right) / 2;
if(c[i]+d[mid]==x)
{
flag = 1;
break;
}
if(c[i]+d[mid]>x)
right = mid - 1;
else
left = mid + 1;
}
if(left==right && c[i]+d[left]==x) flag = 1;
if(flag==1) break;
}
if(flag == 1)
printf("YES\n");
else
printf("NO\n");
}
}
return 0;
}View Code
题意:给出三个数列a,b,c和一组数x,求对于每个x是否存在 ai + bj + ck = x;其中三个数列最多各有500个数,x最多有1000个。
思路: 将数列a,b两两相加,组成一个数列(500*500)。
然后枚举数列c,利用二分从合并数列中查找满足条件的数。
注意事项:二分循环的条件为while(left<right) ,所以若跳出循环后left==right,要加一组判断。if(left==right && c[i]+d[left]==x) flag = 1;
1 #include <cstdio>
2 #include <cstring>
3 #include <algorithm>
4 using namespace std;
5 #define N 505
6 #define S 1005
7
8 int a
,b
,c
,d[N*N];
9
int main()
{
int l, n, m, s, x, Case = 0;
while(scanf("%d%d%d",&l,&n,&m)!=EOF)
{
printf("Case %d:\n",++Case);
for(int i=1; i<=l; i++)
scanf("%d",&a[i]);
for(int i=1; i<=n; i++)
scanf("%d",&b[i]);
for(int i=1; i<=m; i++)
scanf("%d",&c[i]);
int num = 0;
for(int i=1; i<=l; i++)
{
for(int j=1; j<=n; j++)
d[++num] = a[i] + b[j];
}
sort(d+1,d+num+1);
scanf("%d",&s);
while(s--)
{
scanf("%d",&x);
int flag = 0;
for(int i=1; i<=m; i++)
{
int left = 1, right = num, mid;
while(left<right)
{
mid = (left + right) / 2;
if(c[i]+d[mid]==x)
{
flag = 1;
break;
}
if(c[i]+d[mid]>x)
right = mid - 1;
else
left = mid + 1;
}
if(left==right && c[i]+d[left]==x) flag = 1;
if(flag==1) break;
}
if(flag == 1)
printf("YES\n");
else
printf("NO\n");
}
}
return 0;
}View Code
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