test Week 5

来源：互联网发布：手机app界面设计软件编辑：程序博客网时间：2024/05/19 04:04

180min 2/5 出了三题的主体，但每一道都WA了一下，因各种因素而小卡，还要要多积累

A题，a[i] = a[i-1] + a[-2] ，关系式能出，赞一个！但是没用longlong WA了一次

#include <iostream>using namespace std;const int MAXN = 45 + 5;int N;long long a[MAXN];int main(){a[1] = 2;a[2] = 2;cin >> N;for (int i = 3; i <= N; i++) {a[i] = a[i-1] + a[i-2];}cout << a[N] << endl;}

B题

状态转移方程： d[i][j] = min{d[i-1][j-1] + 141, d[i-1][j] + 100, d[i][j-1] +100}, 但d[i-1][j-1]需合法。

double的四舍五入WA了一次，之前一直以为ceil能做四舍五入，ceil实际上是向上取整数，%.0f不赖

#include <iostream>#include <cstdio>#include <cmath>using namespace std;const int MAXN = 1000 + 5;const int INF = 300000;double dist[MAXN][MAXN];bool hasd[MAXN][MAXN];int N, M, K, a, b;int main(){cin >> N >> M >> K;for (int i = 0; i < K; i++) {scanf("%d %d", &a, &b);hasd[a][b] = true;}for (int i = 0; i <= N; i++) {for (int j = 0; j <= M; j++) {dist[i][j] = INF;}}dist[0][0] = 0;for (int i = 0; i <= N; i++) {for (int j = 0; j <= M; j++) {if (dist[i][j] + 100 < dist[i][j+1]) {dist[i][j+1] = dist[i][j] + 100;}if (dist[i][j] + 100 < dist[i+1][j]) {dist[i+1][j] = dist[i][j] + 100;}if (hasd[i+1][j+1] && dist[i][j]+ 100*sqrt(2) < dist[i+1][j+1]) {dist[i+1][j+1] = dist[i][j]+100*sqrt(2);}}}/*for (int j = M; j >= 0 ; j--) {for (int i = 0; i <= N; i++) {cout << dist[i][j] <<  ' ';}cout << endl;}*/printf("%0.f\n", dist[N][M] + 0.0001);}

C也好找状态转移：

令dp[i][j]=把前i头马放在前j个马厩里的最小值，那么答案是dp[n][k]

dp[i][j] = min{ dp[i-1][j-1] , dp[i-2][j-1] + A1, dp[i-3][j-1] + A2 ……}

这个状态转移方程的意思是dp[i][j]的最小值取决于dp[i-1][j-1], dp[i-2][j-1] ……把马放前j-1个马厩，后面的吗放第j个马厩

万万没想到，计算A1、A2这些值成了时间瓶颈TLE。。。。原来的扎做法是几下第i匹马是黑的还是白的然后就扫一遍，后来换成几下前i匹马有x头黑的， O(1)才过。。。

#include <iostream>#include <cstdio>using namespace std;const int MAXN = 500 + 5;const int INF = 80000;int N, K, tmp;int black[MAXN];bool isb[MAXN];int dp[MAXN][MAXN];int test(int st, int ed) {int bl = black[ed]-black[st]+isb[st];int white = ed-st+1-bl;return bl*white;}int main() {scanf("%d%d", &N, &K);int count = 1;for (int i = 1; i <= N; i++) {scanf("%d", &tmp);if (tmp == 1) {isb[i] = true;count++;}black[i] = count;}for (int i = 1; i <= N; i++) {for (int j = 1; j <= K; j++) {dp[i][j] = INF;}}for (int i = 1; i <= N-K+1; i++) {dp[i][1] = test(1, i);}/*for (int i = 1; i <= N; i++) {cout << black[i] << endl;}for (int i = 1; i <= N; i++) {for (int j = i; j <= N; j++) {cout << i << ' ' << j << endl;cout << test(i, j) << endl;}}*/for (int i = 2; i <= K; i++) {dp[i][i] = 0;for (int j = i+1; j <= N-(K-i); j++) {for (int q = i-1; q < j; q++) {if (dp[j][i] > dp[q][i-1] + test(q+1,j)) {dp[j][i] = dp[q][i-1] + test(q+1,j);}}}}cout << dp[N][K] << endl;}

0 0