UVA 467 - Synching Signals（数论）

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UVA 467 - Synching Signals

题目链接

题意：给定几个红绿灯，每个红绿灯time表示，time秒红灯，time - 5秒绿灯， 5秒黄灯.
然后求全部灯变绿之后，在一次有灯变换之后，全是绿灯需要的时间

思路：由于只要算1小时，也就是3600秒，直接暴力过去，每次时间加上当前最小能变换灯的时间，然后记录一下每个灯的颜色状态，直到全变绿为止

代码：

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;char str[105];int ti[15], s[15], r[15], n;int cas = 0;void solve() {int k = ti[0];//s = 0绿， s = 1黄, s = 2红 memset(s, 0, sizeof(s));memset(r, 0, sizeof(r));for (int i = 1; i < n; i++)k = min(k, ti[i]);k -= 5; int t = k;for (int i = 0; i < n; i++) {r[i] = (ti[i] - 5) - k;if (r[i] == 0) {r[i] = 5;s[i] = 1;  }  else s[i] = 0;} while (t <= 3600) { k = r[0]; int sum = 0; for (;sum < n; sum++) if (s[sum]) break;if (sum == n) break; for (int i = 1; i < n; i++) k = min(k, r[i]);for (int i = 0; i < n; i++) {r[i] -= k;if (r[i]) continue;if (s[i] == 0) {s[i] = 1;r[i] = 5;   }   else if (s[i] == 1) {   s[i] = 2;   r[i] = ti[i];      }      else {      s[i] = 0;      r[i] = ti[i] - 5;   }  }  t += k;  }  if (t > 3600) printf("Set %d is unable to synch after one hour.\n", ++cas);  else {  int minu = t / 60;  int sec = t % 60;  printf("Set %d synchs again at %d minute(s) and %d second(s) after all turning green.\n", ++cas, minu, sec);   }}int main() {while (gets(str) != NULL) {int len = strlen(str); n = 0; str[len++] = ' ';int num = 0;  for (int i = 0; i < len; i++) {  if (str[i] >= '0' && str[i] <= '9') {  num = num * 10 + str[i] - '0';  continue;        }        ti[n++] = num;        num = 0;  }  solve(); }return 0;}

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