题意：有n门考试，对于考试i，不复习它能考si分，复习它的每小时能提高ai分，每过一小时ai会减小di，也就是说，连续复习某门科目每小时提高的分为ai, ai-di, ai-2di...，但每门考试最高分不超过100分，给定每门考试的考试时间，且考试本身不占时间，合理安排复习计划，问能否让所有考试都不低于60分，如果可以，总和最大为多少。
思路：比较明显的贪心，但难以想到正确的贪心策略。我们把问题分成两步，所有考试全部通过60分和得分总和最大。对于第一步，对于不复习分数低于60分的科目，先复习一段时间让它分数达到60分，贪心从考试时间往前延伸找空闲时间复习（一旦找不到空闲时间，则无解了）。对于第二步，按时间顺序的逆序处理，对于某个时间t，枚举所有考试时间大于等于t的科目，找到某一科目，使得复习它提高的分数最高，那么时间t就用来复习这门功课。
1 #pragma comment(linker, "/STACK:10240000,10240000") 2 3 #include <iostream> 4 #include <cstdio> 5 #include <algorithm> 6 #include <cstdlib> 7 #include <cstring> 8 #include <map> 9 #include <queue> 10 #include <deque> 11 #include <cmath> 12 #include <vector> 13 #include <ctime> 14 #include <cctype> 15 #include <set> 16 #include <bitset> 17 #include <functional> 18 #include <numeric> 19 #include <stdexcept> 20 #include <utility> 21 22 using namespace std; 23 24 #define mem0(a) memset(a, 0, sizeof(a)) 25 #define mem_1(a) memset(a, -1, sizeof(a)) 26 #define lson l, m, rt << 1 27 #define rson m + 1, r, rt << 1 | 1 28 #define define_m int m = (l + r) >> 1 29 #define rep_up0(a, b) for (int a = 0; a < (b); a++) 30 #define rep_up1(a, b) for (int a = 1; a <= (b); a++) 31 #define rep_down0(a, b) for (int a = b - 1; a >= 0; a--) 32 #define rep_down1(a, b) for (int a = b; a > 0; a--) 33 #define all(a) (a).begin(), (a).end() 34 #define lowbit(x) ((x) & (-(x))) 35 #define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {} 36 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {} 37 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {} 38 #define pchr(a) putchar(a) 39 #define pstr(a) printf("%s", a) 40 #define sstr(a) scanf("%s", a) 41 #define sint(a) scanf("%d", &a) 42 #define sint2(a, b) scanf("%d%d", &a, &b) 43 #define sint3(a, b, c) scanf("%d%d%d", &a, &b, &c) 44 #define pint(a) printf("%d ", a) 45 #define test_print1(a) cout << "var1 = " << a << endl 46 #define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl 47 #define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = " << b << ", var3 = " << c << endl 48 49 typedef long long LL; 50 typedef pair<int, int> pii; 51 typedef vector<int> vi; 52 53 const int dx[8] = {0, 0, -1, 1, 1, 1, -1, -1}; 54 const int dy[8] = {-1, 1, 0, 0, 1, -1, 1, -1 }; 55 const int maxn = 3e4 + 7; 56 const int md = 10007; 57 const int inf = 1e9 + 7; 58 const LL inf_L = 1e18 + 7; 59 const double pi = acos(-1.0); 60 const double eps = 1e-4; 61 62 template<class T>T gcd(T a, T b){return b==0?a:gcd(b,a%b);} 63 template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;} 64 template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;} 65 template<class T>T condition(bool f, T a, T b){return f?a:b;} 66 template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];} 67 int make_id(int x, int y, int n) { return x * n + y; } 68 69 struct Node { 70 int r, s, a, d; 71 Node(int r = 0, int s = 0, int a = 0, int d = 0): r(r), s(s), a(a), d(d) {} 72 bool operator < (const Node that) const { 73 return r < that.r; 74 } 75 }; 76 Node node[100]; 77 bool occ[1000]; 78 79 int main() { 80 //freopen("in.txt", "r", stdin); 81 int n; 82 while (cin >> n) { 83 int maxt = 0; 84 bool ok = true; 85 mem0(occ); 86 rep_up0(i, n) { 87 int s, t, a, d, tmp = 0; 88 cin >> s >> t >> a >> d; 89 node[i] = Node(t, s, a, d); 90 max_update(maxt, t); 91 } 92 sort(node, node + n); 93 rep_up0(i, n) { 94 int cur = node[i].r; 95 while (node[i].s < 60) { 96 while (occ[cur] && cur) cur--; 97 if (!cur || node[i].a <= 0) { 98 ok = false; 99 break; 100 } 101 occ[cur] = true; 102 node[i].s += node[i].a; 103 node[i].a -= node[i].d; 104 } 105 if (!ok) break; 106 } 107 rep_down1(i, maxt) { 108 if (!occ[i]) { 109 int s = 0, p; 110 rep_up0(j, n) { 111 if (node[j].r >= i) { 112 int dat = node[j].a; 113 if (node[j].s + dat > 100) dat = 100 - node[j].s; 114 if (dat > s) { 115 s = dat; 116 p = j; 117 } 118 } 119 } 120 if (s) { 121 node[p].s += s; 122 node[p].a -= node[p].d; 123 } 124 } 125 } 126 int ans = 0; 127 rep_up0(i, n) { 128 ans += node[i].s; 129 } 130 if (ok) cout << ans << endl; 131 else puts("you are unlucky"); 132 } 133 }