1967 - 保险丝
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。
Input
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。
Output
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。
Examples
Input
2 2 10 5 7 1 2 3 6 10 2 5 7 2 1 2 3 1 3 0 0 0
Output
Sequence 1 Fuse was blown. Sequence 2 Fuse was not blown. Maximal power consumption was 9 amperes.
Solution C
#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 //
Solution C++
#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; }