Codeforces Round #325 (Div. 2)ABC
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A - Alena's Schedule
分析:遇到1答案加一,遇到101加一。
#include <iostream>using namespace std;int main(){ int n, ans = 0, a[105]; cin >> n; for(int i = 0 ; i < n ; i ++) { cin >> a[i]; ans += a[i]; if(i > 1&&a[i-1] == 0&&a[i-2] == 1&&a[i] == 1) ans ++; } cout << ans << endl; return 0;}B - Laurenty and Shop
分析:用数组a把第一排房子到第一个房子的距离存下来,再用一个数组c吧第二排房子和最后一个房子的距离存下来,数组c表示第一排和第二排房子之间的距离,则取a+b+c的最小值就是第一个到最后一个房子之间的距离。返回时不走一样的路,把去超市时的c[i]标记一下,下次不走这条路就行了,a+b+c还是取最小值,相加就可以了。
#include <iostream>#include <cstring>using namespace std;const int INF = 999999999;int main(){ int n, t = INF, t2 = INF, j; int a[2][100], b[100]; memset(a, 0, sizeof(a)); cin >> n; for(int i = 1 ; i < n ; i ++) cin >> a[0][i]; for(int i = 0 ; i < n-1 ; i ++) cin >> a[1][i]; for(int i = 0 ; i < n ; i ++) cin >> b[i]; for(int i = 1 ; i < n ; i ++) a[0][i] += a[0][i-1]; for(int i = n-2 ; i >= 0 ; i --) a[1][i] += a[1][i+1]; for(int i = 0 ; i < n ; i ++) { if(a[0][i]+a[1][i]+b[i] < t) { t = a[0][i] + a[1][i] + b[i]; j = i; } } b[j] = INF; for(int i = 0 ; i < n ; i ++) if(a[0][i]+a[1][i]+b[i] < t2) t2 = a[0][i] + a[1][i] + b[i]; cout << t+t2 << endl; return 0;}C - Gennady the Dentist
分析:模拟。
#include <iostream>#include <cstring>#define LL long longusing namespace std;const int maxn = 4000 + 5;const int INF = 999999999;LL n, b, ans;LL v[maxn], d[maxn], p[maxn], q[maxn], a[maxn];int main(){ ans = 0; cin >> n; for(int i = 1 ; i <= n ; i ++) { cin >> v[i] >> d[i] >> p[i]; q[i] = p[i]; } for(int i = 1 ; i <= n ; i ++) { if(p[i] >= 0) { a[ans++] = i; b = 0; for(int j = i+1 ; j <= n ; j ++) if(p[j] >= 0) { p[j] -= v[i] + b; if(p[j] < 0) b += d[j]; if(v[i]) v[i]--; } } } cout << ans << endl; for(int i = 0 ; i < ans ; i ++) cout << a[i] << ' '; cout << endl; return 0;}
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