Cut Pieces
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)Total Submission(s): 69 Accepted Submission(s): 27
Consider one way to color the blocks. We call a consecutive sequence of blocks with the same color a "piece". For example, sequence "Yellow Yellow Red" has two pieces and sequence "Yellow Red Blue Blue Yellow" has four pieces. What is S, the total number of pieces of all possible ways to color the blocks?
This is not your task. Your task is to permute the blocks (together with its corresponding ai) so that S is maximized.
Following are 2*T lines. For every two lines, the first line is n, length of sequence; the second line contains n numbers, a1, ..., an.
Sum of all n <= 106.
All numbers in the input are positive integers no larger than 109.
Each line contains one number, the answer to the corresponding test case.
Since the answers can be very large, you should output them modulo 109+7.
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <vector>
#include <cmath>
using namespace std;
//
const int V = 1000000 + 50;
const int MaxN = 80 + 5;
const int mod = 1000000000 + 7;
const __int64 INF = 0x7FFFFFFFFFFFFFFFLL;
const int inf = 0x7fffffff;
int T, n, num[V], ans[V];
__int64 dp[V], sum[V];
int main() {
int i, j;
scanf("%d", &T);
while(T--) {
scanf("%d", &n);
for(i = 0; i < n; ++i)
scanf("%d", &num[i]);
sort(num, num + n);
int ii = 0, jj = n - 1;
for(i = 0; i < n; ++i) {
if(i % 2 == 0) {
ans[i] = num[ii];
ii++;
}
else {
ans[i] = num[jj];
jj--;
}
}
sum[n] = 1;
sum[n - 1] = ans[n - 1];
dp[n - 1] = ans[n - 1];
for(i = n - 2; i >= 0; --i) {
if(ans[i] >= ans[i + 1])
dp[i] = (ans[i] - ans[i + 1]) * (dp[i + 1] + sum[i + 1]) + (sum[i + 1] - sum[i + 2] + dp[i + 1]) * ans[i + 1];
else
dp[i] = ans[i] * (dp[i + 1] + sum[i + 1] - sum[i + 2]);
dp[i] %= mod;
dp[i] = (dp[i] + mod) % mod;
sum[i] = (sum[i + 1] * ans[i]) % mod;
}
printf("%d\n", dp[0]);
}
}