hdu 4655 Cut Pieces（DP 贪心）

根据题意先求出dp公式：

if a[i-1] < a[i] then

dp[i] = (dp[i-1]*a[i]) + a[0]*a[1]*....a[i-1]*(a[i] - 1);

else

dp[i] = (dp[i-1]*a[i]) + a[0]*a[1]*....a[i-2]*(a[i-1]-1)*a[i];

可见相邻的数差值越大所能造成的影响因素越大，那么基于贪心的原则可得到变化 诸如：1,2,3,4,5 --> 1,5,2,4,3

#include <cstdio>
#include <algorithm>
using namespace std;
const int MAXN = 1000005;
typedef __int64 ll;
ll a[MAXN], dp[MAXN], b[MAXN];
const ll mod = 1000000007;
int n;
void solve()
{
dp[0] = a[0];
ll alio2 = 1, alio1 = a[0];
for (int i = 1; i< n; ++i)
{
if (a[i] > a[i-1])
{
dp[i] = (dp[i-1]*a[i]%mod + alio1*(a[i]-1)%mod)%mod;
}
else
{
dp[i] = (dp[i-1]*a[i]%mod + alio2*(a[i-1]-1)%mod*a[i]%mod)%mod;
}
alio2 = alio1;
alio1 = alio1 * a[i] % mod;
}
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
#endif
int t;
scanf("%d", &t);
while (t--)
{
scanf("%d", &n);
for (int i = 0; i< n; ++i) scanf("%I64d", b+i);
sort(b, b+n);
for (int i = 0, j = 0; j< n; i++, j+=2)
{
a[j] = b[i];
}
for (int i = n-1, j = 1; j<n; i--, j+=2)
{
a[j] = b[i];
}
solve();
printf("%I64d\n", dp[n-1]);
}
return 0;
}