2个 spa ble log pri scan string 需要 tar
题目链接:http://acm.split.hdu.edu.cn/showproblem.php?pid=5037
题解:为了让放的石头有意义肯定是没l+1的距离放2个也就是说假设现在位置为pos那么在pos+1放一个在pos+l+1放一个这样就需要跳两次。于是这题要考虑的就是当前位置和前一个石头放的位置因为如果前一个石头的位置到a[i]的距离小于等于l那么当前位置就不能被走到。所以就拿l+1的距离来贪心。
- #include < iostream > #include < cstring > #include < cstdio > #include < algorithm > using namespace std;
- const int M = 2e5 + 10;
- int a[M];
- int main() {
- int t,
- n,
- m,
- l;
- scanf("%d", &t);
- int Case = 0;
- while (t--) {
- scanf("%d%d%d", &n, &m, &l);
- for (int i = 0; i < n; i++) {
- scanf("%d", &a[i]);
- }
- sort(a, a + n);
- a[n] = m;
- int pos = 0,
- pre = -l;
- int ans = 0;
- for (int i = 0; i <= n; i++) {
- ans += (a[i] - pos) / (l + 1) * 2;
- pre += (a[i] - pos) / (l + 1) * (l + 1);
- if (a[i] - pre > l) {
- pre = pos + (a[i] - pos) / (l + 1) * (l + 1);
- pos = a[i];
- ans++;
- } else {
- pos = a[i];
- }
- }
- printf("Case #%d: %d\n", ++Case, ans);
- }
- return 0;
- }
hdu 5037 Frog(贪心)
来源: http://www.bubuko.com/infodetail-2357728.html