HDU-4972-A Simple Dynamic Programming Problem (Mathematics + dp)

Question: In an NBA game, the two sides scored n times (n <= 100000). Whichever side, they scored one goal (the score can only be 1, 2, 3 ), record the score once (the absolute value of the score difference between the two teams) and ask the number of scores of the last two teams.

Question link: http://acm.hdu.edu.cn/showproblem.php? PID = 1, 4972

--> How can I obtain the score if I know the last score is k? If the final score of one party with a lower score is X, the final score of the other party is x + K, and the final total score of both parties is S, then x = (S-k)/2, the score can be obtained from both parties .. Therefore, as long as you know the sum of the total score of both parties, you can determine the score of both parties .. As a result, the question is transformed into the sum of the total points and quantity of the two sides.

After knowing the sum of the Double score, you can find X: x + k and x + K: X. Therefore, if X! = X + k is k! When the value is 0, the final result is multiplied by 2 ..

Status: DP [I] indicates the sum of the total scores of both sides after the first goal.

State transition equation: DP [I] = DP [I-1] + 1 (when (diff [I] = 1 & diff [I-1] = 2) | (when diff [I] = 2 & diff [I-1] = 1)

Border: DP [0] = 1, diff [0] = 0

Note: If the last score difference is 2 and the score difference is 1, the loser scores 1 or 3, and two total scores are produced .. The last score difference is 1, and the current score difference is 2 ..

Submit g ++ and use ABS (INT). Ce (check the cmath in mingw. ABS does not provide the int parameter). c ++ won't (go over msdn, the provided parameter has an int value ).. I always use ABS to calculate the absolute value of an integer, and use fabs to calculate the absolute value of a floating point ..

#include <cstdio>#include <cmath>using std::abs;int main(){    int T, N, diff, kase = 0;    scanf("%d", &T);    while (T--)    {        int dp = 1, last = 0;        bool ok = true;        scanf("%d", &N);        while (N--)        {            scanf("%d", &diff);            if (!ok) continue;            if (abs(diff - last) > 3 || (diff == last && diff != 1))            {                ok = false;            }            else            {                if ((diff == 2 && last == 1) || (diff == 1 && last == 2))                {                    dp++;                }                last = diff;            }        }        if (diff != 0)        {            dp <<= 1;        }        printf("Case #%d: ", ++kase);        ok ? printf("%d\n", dp) : puts("0");    }    return 0;}

HDU-4972-A Simple Dynamic Programming Problem (Mathematics + dp)