【meet in the middle】的几个例题【BZOJ 4800】&【SPOJ ABCDEF】

【题目】SPOJABCDEF

传送门

【Solution】

　　可以将原式子转化成ab+c=d(e+f)，用O(n3)来枚举，统计时自己手写一个二分查找（同map果断TLE）

　　还有，在枚举d时要保证d不为0，否则会WA，不知道为什么。

【Code】

#include<cstdio>
#include<cstring>
#include<algorithm>

#define N 110
#define rep(i,x,y) for(int i = x;i <= y;i++)
using namespace std;
int s
,cnt,num[N*N*N];

int high(int x)
{
int L = 1,R = cnt,res = 0;
while(L <= R)
{
int mid = (L+R)>>1;
if(x >= num[mid]){L = mid+1;res = mid;}
else R = mid - 1;
}
return res;
}

int low(int x)
{
int L = 1,R = cnt,res = 0;
while(L <= R)
{
int mid = (L+R)>>1;
if(x <= num[mid]){R = mid-1;res = mid;}
else L = mid + 1;
}
return res;
}

int main()
{
int n,ans = 0;
scanf("%d",&n);
for(int i = 1;i <= n;i++) scanf("%d",&s[i]);
cnt = 0;
rep(a,1,n) rep(b,1,n) rep(c,1,n)
num[++cnt] = s[a]*s[b]+s[c];
sort(num+1,num+cnt+1);

rep(d,1,n) rep(e,1,n) rep(f,1,n)
if(s[d] != 0) {
int x = s[d]*(s[e]+s[f]);
int r = high(x),l = low(x);
if(r >= l && r != 0 && l != 0) ans += r-l+1;
}
printf("%lld\n",ans);
return 0;
}

【题目】BZOJ 4800

有n个物品，m块钱，给定每个物品的价格，求买物品的方案数。

【Solution】

　　朴素的折半搜索+剪枝

【Code】

#include<cstdio>
#include<cstring>
#include<algorithm>

#define N 45
typedef long long ll;
using namespace std;
ll a
,num1[1100000],num2[1100000],m;
<
4000
span class="hljs-keyword">int deep,cnt1,cnt2,n;

void dfs1(ll sum,int d)
{
if(d == deep+1) {num1[++cnt1] = sum;return;}
if(sum+a[d] <= m) dfs1(sum+a[d],d+1);
dfs1(sum,d+1);
}

void dfs2(ll sum,int d)
{
if(d == n+1) {num2[++cnt2] = sum;return;}
if(sum+a[d] <= m) dfs2(sum+a[d],d+1);
dfs2(sum,d+1);
}

int main()
{
scanf("%d%lld",&n,&m);
for(int i = 1;i <= n;i++) scanf("%lld",&a[i]);
if(n == 1) {
if(a[1] <= m) puts("2"); else puts("1");
return 0;
}
deep = n/2;
cnt1 = cnt2 = 0;
dfs1(a[1],2); dfs1(0,2);
sort(num1+1,num1+cnt1+1);

deep++;
dfs2(a[deep],deep+1); dfs2(0,deep+1);
sort(num2+1,num2+cnt2+1);

ll pos = cnt1,ans = 0;
for(int i = 1;i <= cnt2;i++) {
while(num2[i] > m - num1[pos] && pos > 0) pos--;
ans += pos;
}
printf("%lld\n",ans);
return 0;
}