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uva 580 - Critical Mass(dp)

题目链接:uva 580 - Critical Mass

题目大意:给定一个栈,向栈里连续添加n次,每次可以添加U或者L,如果出现连续三个U则为不安全,问有多少种不安全的可能。

解题思路:先求出安全的,用总数减去安全的即为答案。
dp[i][j]表示以第i个位置结尾时,有末尾有j个连续的U。
还有一种解法,dp[i]表示第i个位置以L结尾的总数,dp[i] = dp[i-1] + dp[i-2] + dp[i-3];

解法一:
#include <cstdio>
#include <cstring>

typedef long long ll;
ll dp[40][3];

ll solve (int n) {
    memset(dp, 0, sizeof(dp));
    dp[0][0] = 1;

    for (int i = 1; i <= n; i++) {

        for (int j = 0; j < 3; j++)
            dp[i][0] += dp[i-1][j];

        for (int j = 1; j <= 2; j++)
            dp[i][j] = dp[i-1][j-1];
    }
    ll ans = 0;
    for (int i = 0; i < 3; i++)
        ans += dp[n][i];
    return (1LL<<n)-ans;
}

int main () {
    int n;
    while (scanf("%d", &n) == 1 && n) {
        printf("%lld\n", solve(n));
    }
    return 0;
}
解法二:
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
typedef long long ll;
ll dp[40];

ll solve(int n) {

    memset(dp, 0, sizeof(dp));
    dp[0] = dp[1] = 1;
    for (int i = 2; i <= n; i++) {

        for (int j = 1; j <= min(3, i); j++)
            dp[i] += dp[i-j];
    }

    ll ans = 0;
    for (int i = 0; i <= min(n, 2); i++)
        ans += dp[n-i];
    return (1LL<<n)-ans;
}

int main () {
    int n;
    while (scanf("%d", &n) == 1 && n) {
        printf("%lld\n", solve(n));
    }
    return 0;
}