UVA 467 - Synching Signals（数论）

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;
}