codeforces 788A——Functions again(动态规划)
2017-03-30 20:12
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A. Functions again
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Something happened in Uzhlyandia again... There are riots on the streets... Famous Uzhlyandian superheroes Shean the Sheep and Stas the Giraffe were called in order to save the situation. Upon the arriving, they found that citizens are worried about maximum
values of the Main Uzhlyandian Function f, which is defined as follows:
In the above formula, 1 ≤ l < r ≤ n must hold, where n is
the size of the Main Uzhlyandian Array a, and |x| means
absolute value of x. But the heroes skipped their math lessons in school, so they asked you for help. Help them calculate the maximum
value of f among all possible values of l and r for
the given array a.
Input
The first line contains single integer n (2 ≤ n ≤ 105) —
the size of the array a.
The second line contains n integers a1, a2, ..., an (-109 ≤ ai ≤ 109) —
the array elements.
Output
Print the only integer — the maximum value of f.
Examples
input
output
input
output
Note
In the first sample case, the optimal value of f is reached on intervals [1, 2] and [2, 5].
In the second case maximal value of f is reachable only on the whole array.
简单dp,开二维dp。因为结果子段一定是正负交替的。
dp[i][0]表示当前的差值取正,dp[i][1]表示当前的差值取负
#define _CRT_SECURE_NO_WARNINGS
#include<cstdio>
#include<cstring>
#include<queue>
#include <cstdlib>
#include<map>
#include <set>
#include <queue>
using namespace std;
const int MAXN = 100010;
long long INF = 0x7fffffffffffffff;
long long a[MAXN];
long long dp[MAXN][2];
long long d[MAXN];
int n;
int main()
{
int n;
scanf("%d", &n);
long long MAX = -INF;
for (int i = 1; i <= n; i++) {
scanf("%I64d", &a[i]);
if (i > 1) {
d[i] = max(a[i], a[i - 1]) - min(a[i], a[i - 1]);
}
}
for (int i = 2; i <= n; i++) {
dp[i][0] = max(d[i], dp[i - 1][1]+d[i]);
dp[i][1] = max(-d[i], dp[i - 1][0] - d[i]);
MAX = max(dp[i][0], MAX);
MAX = max(dp[i][1], MAX);
}
printf("%I64d\n", MAX);
//system("pause");
return 0;
}
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Something happened in Uzhlyandia again... There are riots on the streets... Famous Uzhlyandian superheroes Shean the Sheep and Stas the Giraffe were called in order to save the situation. Upon the arriving, they found that citizens are worried about maximum
values of the Main Uzhlyandian Function f, which is defined as follows:
In the above formula, 1 ≤ l < r ≤ n must hold, where n is
the size of the Main Uzhlyandian Array a, and |x| means
absolute value of x. But the heroes skipped their math lessons in school, so they asked you for help. Help them calculate the maximum
value of f among all possible values of l and r for
the given array a.
Input
The first line contains single integer n (2 ≤ n ≤ 105) —
the size of the array a.
The second line contains n integers a1, a2, ..., an (-109 ≤ ai ≤ 109) —
the array elements.
Output
Print the only integer — the maximum value of f.
Examples
input
5 1 4 2 3 1
output
3
input
4 1 5 4 7
output
6
Note
In the first sample case, the optimal value of f is reached on intervals [1, 2] and [2, 5].
In the second case maximal value of f is reachable only on the whole array.
简单dp,开二维dp。因为结果子段一定是正负交替的。
dp[i][0]表示当前的差值取正,dp[i][1]表示当前的差值取负
#define _CRT_SECURE_NO_WARNINGS
#include<cstdio>
#include<cstring>
#include<queue>
#include <cstdlib>
#include<map>
#include <set>
#include <queue>
using namespace std;
const int MAXN = 100010;
long long INF = 0x7fffffffffffffff;
long long a[MAXN];
long long dp[MAXN][2];
long long d[MAXN];
int n;
int main()
{
int n;
scanf("%d", &n);
long long MAX = -INF;
for (int i = 1; i <= n; i++) {
scanf("%I64d", &a[i]);
if (i > 1) {
d[i] = max(a[i], a[i - 1]) - min(a[i], a[i - 1]);
}
}
for (int i = 2; i <= n; i++) {
dp[i][0] = max(d[i], dp[i - 1][1]+d[i]);
dp[i][1] = max(-d[i], dp[i - 1][0] - d[i]);
MAX = max(dp[i][0], MAX);
MAX = max(dp[i][1], MAX);
}
printf("%I64d\n", MAX);
//system("pause");
return 0;
}
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