HDU 4972 A simple dynamic programming problem

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随机输出保平安

#include <cstdio>#include <cmath>#include <iostream>#include <cstring>#include <algorithm>using namespace std;typedef long long ll;const int N = 100005;int a[N];int main() {    int T, cas = 0;    scanf("%d", &T);    while(T-- > 0) {        int n; scanf("%d", &n);        bool flag = 1;        a[0] = 0;        for(int i = 1; i <= n; i ++) {            scanf("%d", &a[i]);            if(abs(a[i] - a[i-1]) > 3) flag = false;            if(a[i] == a[i-1] && a[i] != 1) flag = false;        }        if(!flag) {            printf("Case #%d: 0\n", ++cas);            continue;        }        int ans = 1;        for(int i = 2; i <= n; i ++) {            if(a[i] == 1 && a[i-1] == 2) ans ++;            else if(a[i] == 2 && a[i-1] == 1) ans ++;        }        printf("Case #%d: ", ++cas);        if(a[n] == 0) printf("%d\n", ans);        else printf("%d\n", ans * 2);    }    return 0;}

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