POJ Problem 1745 Divisibility 【dp】
2016-08-08 20:45
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Divisibility
Description
Consider an arbitrary sequence of integers. One can place + or - operators between integers in the sequence, thus deriving different arithmetical expressions that evaluate to different values. Let us, for example, take the sequence: 17, 5, -21, 15. There are
eight possible expressions: 17 + 5 + -21 + 15 = 16
17 + 5 + -21 - 15 = -14
17 + 5 - -21 + 15 = 58
17 + 5 - -21 - 15 = 28
17 - 5 + -21 + 15 = 6
17 - 5 + -21 - 15 = -24
17 - 5 - -21 + 15 = 48
17 - 5 - -21 - 15 = 18
We call the sequence of integers divisible by K if + or - operators can be placed between integers in the sequence in such way that resulting value is divisible by K. In the above example, the sequence is divisible by 7 (17+5+-21-15=-14) but is not divisible
by 5.
You are to write a program that will determine divisibility of sequence of integers.
Input
The first line of the input file contains two integers, N and K (1 <= N <= 10000, 2 <= K <= 100) separated by a space.
The second line contains a sequence of N integers separated by spaces. Each integer is not greater than 10000 by it's absolute value.
Output
Write to the output file the word "Divisible" if given sequence of integers is divisible by K or "Not divisible" if it's not.
Sample Input
Sample Output
Source
Northeastern Europe 1999
思路:
用数组dp[i][j]判断后i个数mod k 的余数是不是k。状态转移方程为dp[i][abs(ar[j]-j)%k] = true(减去后面的数)和dp[i][abs(ar[j]+j)%k] = true(加上后面的数)。
#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#define MAXN 10005
using namespace std;
bool dp[MAXN][105];
int ar[MAXN];
int main() {
int t, n, a, b, k;
while (scanf("%d%d", &n, &k) != EOF) {
memset(dp, false, sizeof(dp));
for (int i = 1; i <= n; i++) {
scanf("%d", &ar[i]);
}
dp
[abs(ar
%k)] = true;
for (int i = n - 1; i >= 1; i--) {
for (int j = 0; j < k; j++) {
if (dp[i+1][j]) {
dp[i][abs(j+ar[i])%k] = true;
dp[i][abs(j-ar[i])%k] = true;
}
}
}
if (dp[1][0]) printf("Divisible\n");
else printf("Not divisible\n");
}
return 0;
}
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 11461 | Accepted: 4100 |
Consider an arbitrary sequence of integers. One can place + or - operators between integers in the sequence, thus deriving different arithmetical expressions that evaluate to different values. Let us, for example, take the sequence: 17, 5, -21, 15. There are
eight possible expressions: 17 + 5 + -21 + 15 = 16
17 + 5 + -21 - 15 = -14
17 + 5 - -21 + 15 = 58
17 + 5 - -21 - 15 = 28
17 - 5 + -21 + 15 = 6
17 - 5 + -21 - 15 = -24
17 - 5 - -21 + 15 = 48
17 - 5 - -21 - 15 = 18
We call the sequence of integers divisible by K if + or - operators can be placed between integers in the sequence in such way that resulting value is divisible by K. In the above example, the sequence is divisible by 7 (17+5+-21-15=-14) but is not divisible
by 5.
You are to write a program that will determine divisibility of sequence of integers.
Input
The first line of the input file contains two integers, N and K (1 <= N <= 10000, 2 <= K <= 100) separated by a space.
The second line contains a sequence of N integers separated by spaces. Each integer is not greater than 10000 by it's absolute value.
Output
Write to the output file the word "Divisible" if given sequence of integers is divisible by K or "Not divisible" if it's not.
Sample Input
4 7 17 5 -21 15
Sample Output
Divisible
Source
Northeastern Europe 1999
思路:
用数组dp[i][j]判断后i个数mod k 的余数是不是k。状态转移方程为dp[i][abs(ar[j]-j)%k] = true(减去后面的数)和dp[i][abs(ar[j]+j)%k] = true(加上后面的数)。
#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#define MAXN 10005
using namespace std;
bool dp[MAXN][105];
int ar[MAXN];
int main() {
int t, n, a, b, k;
while (scanf("%d%d", &n, &k) != EOF) {
memset(dp, false, sizeof(dp));
for (int i = 1; i <= n; i++) {
scanf("%d", &ar[i]);
}
dp
[abs(ar
%k)] = true;
for (int i = n - 1; i >= 1; i--) {
for (int j = 0; j < k; j++) {
if (dp[i+1][j]) {
dp[i][abs(j+ar[i])%k] = true;
dp[i][abs(j-ar[i])%k] = true;
}
}
}
if (dp[1][0]) printf("Divisible\n");
else printf("Not divisible\n");
}
return 0;
}
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