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1967 - 保险丝

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时间限制 : 1 秒 内存限制 : 32 MB

现有n个设备，标号分别从1到n，第i个设备的用电量为c[i]。我们有一保险丝，当当前设备使用电量之和超过保险丝的总容量时则会跳闸。接下来有m次操作，初始化所有设备都为关闭状态，每操作一次则改变状态，若当前状态为开启，那么操作完毕后则变为关闭，反之亦然。

题目输入

测试数据有多组。

输入第一行有3个整数，n、m、c，n(n<=20)表示设备的个数，c表示保险丝的总容量。

接下来有n行，每行一个数字，第i行表示第i个设备的容量。    接下来有m行，每行一个数字k，表示掰动第k个设备的开关。

当n、m、c都为0时，输入结束。

题目输出

对于每个样例，先输出样例号（见样例），若跳闸了则输出“Fuse was blown.”（引号不输出）。否则先输出“Fuse was not blown.”，然后另起一行输出在进行设备开关的过程中最大的用电量。

具体格式参见样例。

每个样例的最后输出一个空行。

输入/输出样例

输入格式

2 2 10
5
7
1
2
3 6 10
2
5
7
2
1
2
3
1
3
0 0 0

输出格式

Sequence 1
Fuse was blown.

Sequence 2
Fuse was not blown.
Maximal power consumption was 9 amperes.

C语言解答

#include <string.h>
#include <stdio.h>

int a[102];
 
int main()
{
	int n,m,c,max,i,j,f,p,res,t;
	int a[100],o[100];

	t=0;
	while (1==1)
	{
		t++;
		scanf("%d%d%d",&n,&m,&c);

		if (n==0) break;
	
		for (i=1;i<=n;i++)
		{
			scanf("%d",&a[i]);
		}

		max=0;
		memset(o,0,400);
		res=0;

		for (i=1;i<=m;i++)
		{
			scanf("%d",&p);
			if (o[p]==0) o[p]=1; else o[p]=0; 

			if (o[p]==1) 
			{
				res=res+a[p];
				if (res>max) max=res;
			}else res-=a[p];
		}

		printf("Sequence %d\n",t);
	    if (max>c) {printf("Fuse was blown.\n\n"); continue;}

		printf("Fuse was not blown.\n");
		printf("Maximal power consumption was %d amperes.\n\n",max);

	}

	return 0;
}
//0 0 1 1 0 0 0 1
//

C++解答

#include <cstdio>
#include <vector>
using namespace std;

void solve(int n, int m, int tot) {
    vector < int > c(n + 1), op(m);
    vector < int > flag(n + 1, -1);
    for (int i = 1; i <= n; ++i)
        scanf("%d", &c[i]);
    for (int i = 0; i < m; ++i)
        scanf("%d", &op[i]);
    int sum = 0, ans = 0;
    for (int i = 0; i < m; ++i) {
        int k = op[i];
        flag[k] *= -1;
        sum += flag[k] * c[k];
        ans = max(ans, sum);
        if (sum > tot) {
            puts("Fuse was blown.");
            return;
        }
    }
    puts("Fuse was not blown.");
    printf("Maximal power consumption was %d amperes.\n", ans);
}

int main() {
    int n, m, tot, cas = 0;
    while (scanf("%d %d %d", &n, &m, &tot)) {
        if (0 == n && 0 == m && 0 == tot)
            break;
        printf("Sequence %d\n", ++cas);
        solve(n, m, tot);
        puts("");
    }
    return 0;
}

Java解答

import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner scanner = new Scanner(System.in);
		int sequence = 1;
		while (true) {
			int n = scanner.nextInt();
			int m = scanner.nextInt();
			int c = scanner.nextInt();
			if (n == 0 && m == 0 && c == 0) {
				break;
			}
			int[] array = new int[n + 1];
			for (int i = 1; i <= n; i++) {
				array[i] = scanner.nextInt();
			}
			boolean[] array2 = new boolean[n + 1];
			int count = 0;
			boolean flag = false;
			int max = 0;
			for (int i = 0; i < m; i++) {
				int index = scanner.nextInt();
				if (array2[index]) {
					array2[index] = false;
					count -= array[index];
				} else {
					array2[index] = true;
					count += array[index];
					max = Math.max(max, count);
					if (count > c) {
						flag = true;
					}
				}
			}
			System.out.println("Sequence " + sequence);
			if (flag) {
				System.out.println("Fuse was blown.");
			} else {
				System.out.println("Fuse was not blown.");
				System.out.println("Maximal power consumption was " + max
						+ " amperes.");
			}
			System.out.println();
			sequence++;
		}
		scanner.close();
	}
}
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