同餘dp
先驗知識：
餘數的計算公式：c = a -⌊ a/b⌋ * b
其中，⌊ ⌋為向下取整運算子，向下取整運算稱為Floor，用數學符號⌊ ⌋表示
題目：
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.
分析：
dp重點在於找到狀態轉移
dp[i][j]表示加到第i個數時是否能被j整除
/*同餘dp*/
#include <bits/stdc++.h>
#define N 10010
#define M 110
bool dp[N][M];
int main(){
int t, n, k, a;
/*
取模運算和取餘運算（餘數沒有負數）在負數運算上有差別
printf("%d\n", -13%5);
*/
scanf("%d", &t);
while(t --){
memset(dp, 0, sizeof(dp));
scanf("%d%d", &n, &k);
scanf("%d", &a);
a = abs(a) % k;
dp[0][a % k] = dp[0][(k - a) % k] = true;
for(int i = 1; i < n; ++ i){
scanf("%d", &a);
a = abs(a) % k;
for(int j = 0; j < k; ++ j){
if(dp[i - 1][j]){
dp[i][(j + a) % k] = dp[i][(j - a + k) % k] = true;
}
}
}
dp[n - 1][0] ? printf("Divisible\n") : printf("Not divisible\n");
}
return 0;
}