题目链接：https://ac.nowcoder.com/acm/contest/886#question
A：
光速签到。
1 /* basic header */ 2 #include <bits/stdc++.h> 3 /* define */ 4 #define ll long long 5 #define dou double 6 #define pb emplace_back 7 #define mp make_pair 8 #define sot(a,b) sort(a+1,a+1+b) 9 #define rep1(i,a,b) for(int i=a;i<=b;++i) 10 #define rep0(i,a,b) for(int i=a;i<b;++i) 11 #define eps 1e-8 12 #define int_inf 0x3f3f3f3f 13 #define ll_inf 0x7f7f7f7f7f7f7f7f 14 #define lson (curpos<<1) 15 #define rson (curpos<<1|1) 16 /* namespace */ 17 using namespace std; 18 /* header end */ 19 20 int t; 21 const int maxn = 2e3 + 10; 22 23 int main() { 24 scanf("%d", &t); 25 for (int ca = 1; ca <= t; ca++) { 26 printf("Case #%d: ", ca); 27 char s[maxn], t[27]; 28 scanf("%s", s + 1); 29 scanf("%s", t + 1); 30 map<char, int>m; m.clear(); 31 int len = strlen(s + 1), dry = 0, wet = 0, harm = 0; 32 for (int i = 1; i <= len; i++) 33 if (!m.count(s[i])) m[s[i]] = 1; else m[s[i]]++; 34 for (int i = 1; i <= 26; i++) { 35 if (t[i] == 'd') dry += m[(char)(i + 'a' - 1)]; 36 else if (t[i] == 'w') wet += m[(char)(i + 'a' - 1)]; 37 else if (t[i] == 'h') harm += m[(char)(i + 'a' - 1)]; 38 } 39 if (harm * 4 >= len) puts("Harmful"); 40 else if (harm * 10 <= len) puts("Recyclable"); 41 else if (!wet || dry / wet >= 2) puts("Dry"); 42 else puts("Wet"); 43 } 44 return 0; 45 }
B：
一开始认为可以直接找最长的最靠后的0串进行压缩，后来发现有个大坑：首尾和中间同样长度的0对答案贡献不一样大。在中间的连续0压缩后能少1长度。
可以直接枚举删去哪一段0再计算答案，保证正确。
1 /* basic header */ 2 #include <bits/stdc++.h> 3 /* define */ 4 #define ll long long 5 #define dou double 6 #define pb emplace_back 7 #define mp make_pair 8 #define sot(a,b) sort(a+1,a+1+b) 9 #define rep1(i,a,b) for(int i=a;i<=b;++i) 10 #define rep0(i,a,b) for(int i=a;i<b;++i) 11 #define eps 1e-8 12 #define int_inf 0x3f3f3f3f 13 #define ll_inf 0x7f7f7f7f7f7f7f7f 14 #define lson (curpos<<1) 15 #define rson (curpos<<1|1) 16 /* namespace */ 17 using namespace std; 18 /* header end */ 19 20 const int maxn = 10; 21 int a[maxn]; 22 23 int main() { 24 int t; scanf("%d", &t); 25 for (int caseNum = 1; caseNum <= t; caseNum++) { 26 int startPos = 0, maxLen = 0, currLen = 0; 27 for (int i = 1; i <= 8; i++) { 28 a[i] = 0; 29 for (int j = 1; j <= 16; j++) { 30 int x; scanf("%1d", &x); 31 a[i] = a[i] * 2 + x; 32 } 33 if (!a[i]) currLen++; // currLen记录当前连续的全0段个数 34 else { // 如果遇到全1段，维护答案 35 if (currLen >= maxLen && currLen > 1) { 36 startPos = i - currLen; 37 maxLen = currLen; 38 } 39 currLen = 0; 40 } 41 if (i == 8 && currLen > 1) { // 如果全0段延续到末尾 42 if (currLen > maxLen) // 如果是最长，照常处理 43 startPos = i - currLen + 1, maxLen = currLen; 44 if (currLen == maxLen && startPos == 1) // 否则判断开头有长度跟最大长度相同的全0段 45 startPos = i - currLen + 1; 46 } 47 } 48 printf("Case #%d: ", caseNum); 49 if (startPos == 1) printf(":"); 50 for (int i = 1; i <= 8; i++) { 51 if (i == startPos) { 52 printf(":"); i += maxLen; 53 } 54 if (i > 8) puts(""); 55 else printf("%x%c", a[i], ": "[i == 8]); 56 } 57 } 58 return 0; 59 }
D：
这道题很容易想到二分箱子容积，然后就错了。因为这个题箱子容积跟答案没有单调性！
为了保证正确性，这个题最好的方法就是枚举答案。答案下界显然是max(max(vi),sumV/k)，上界则为sumV/k+max(vi)，这样容积的箱子一定能放下所有东西。然后再逐个物品check一次就好了。
1 /* basic header */ 2 #include <bits/stdc++.h> 3 /* define */ 4 #define ll long long 5 #define dou double 6 #define pb emplace_back 7 #define mp make_pair 8 #define sot(a,b) sort(a+1,a+1+b) 9 #define rep1(i,a,b) for(int i=a;i<=b;++i) 10 #define rep0(i,a,b) for(int i=a;i<b;++i) 11 #define eps 1e-8 12 #define int_inf 0x3f3f3f3f 13 #define ll_inf 0x7f7f7f7f7f7f7f7f 14 #define lson (curpos<<1) 15 #define rson (curpos<<1|1) 16 /* namespace */ 17 using namespace std; 18 /* header end */ 19 20 const int maxn = 1e3 + 10; 21 int v[maxn], puted[maxn]; 22 23 int main() { 24 int t; scanf("%d", &t); 25 for (int caseNum = 1; caseNum <= t; caseNum++) { 26 int n, k, sum = 0; scanf("%d%d", &n, &k); 27 for (int i = 1; i <= n; i++) { 28 scanf("%d", &v[i]); sum += v[i]; 29 } 30 sot(v, n); 31 int limV = max(v[n], sum / k); 32 for (int curV = limV;; curV++) { 33 for (int i = 1; i <= n; i++) puted[i] = 0; 34 int cnt = 0; 35 for (int i = 1; i <= k; i++) { 36 int tmpV = curV; 37 for (int j = n; j > 0; j--) { 38 if (!puted[j] && v[j] <= tmpV) { 39 puted[j] = 1; 40 tmpV -= v[j]; 41 cnt++; 42 } 43 } 44 } if (cnt == n) { 45 printf("Case #%d: %d ", caseNum, curV); 46 break; 47 } 48 } 49 } 50 }
G：
一看就是蔡勒公式相关题。通过一些简单性质可以进行剪枝：日份第一位只能是0123，月份第一位只能是0和1，年份第一位不可能为0。而且合法日期+星期五的限制很紧，题解里谈到对于任何一个加密后的日期，只有几万个排列能将其映射到合法日期。故可只对前弱冠个日期求出所有合法的排列之后，再依次检查是否对所有日期合法。注意要对重复日期去重。
顺便这里补充一下蔡勒公式：
对于1752年9月3日前的日期，ans = (d+2*m+3*(m+1)/5+y+y/4+5) % 7；而对于1752年9月3日后的日期，ans = (d + 2*m+3*(m+1)/5+y+y/4-y/100+y/400)%7。这是因为罗马教皇决定在1582年10月4日后使用格利戈里历法，而英国则是在1752年9月3日后才接受使用格利戈里历法。要注意这个点。
计算出来的答案从0到6分别为星期日到星期六。使用时要注意，对于1月和2月的日期，要变为上一年的13月和14月参与计算。
1 /* basic header */ 2 #include <bits/stdc++.h> 3 /* define */ 4 #define ll long long 5 #define dou double 6 #define pb emplace_back 7 #define mp make_pair 8 #define sot(a,b) sort(a+1,a+1+b) 9 #define rep1(i,a,b) for(int i=a;i<=b;++i) 10 #define rep0(i,a,b) for(int i=a;i<b;++i) 11 #define eps 1e-8 12 #define int_inf 0x3f3f3f3f 13 #define ll_inf 0x7f7f7f7f7f7f7f7f 14 #define lson (curpos<<1) 15 #define rson (curpos<<1|1) 16 /* namespace */ 17 using namespace std; 18 /* header end */ 19 20 int a[15], mon[15] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; 21 const int maxn = 1e5 + 10; 22 string s[maxn]; 23 24 int check(int x) { 25 int year = 0, month = a[s[x][5] - 'A'] * 10 + a[s[x][6] - 'A'], day = a[s[x][8] - 'A'] * 10 + a[s[x][9] - 'A']; 26 for (int i = 0; i < 4; i++) 27 year = year * 10 + a[s[x][i] - 'A']; 28 if (year % 4 == 0 && year % 100 || year % 400 == 0) mon[2] = 29; 29 else mon[2] = 28; 30 if (year < 1600 || year > 9999 || month < 1 || month > 12 || day < 1 || day > mon[month]) 31 return 0; 32 if (month <= 2) month += 12, year--; 33 int ans = (day + 2 * month + 3 * (month + 1) / 5 + year + year / 4 - year / 100 + year / 400 + 1) % 7; 34 if (ans == 5) return true; 35 else return false; 36 } 37 38 int main() { 39 int t; scanf("%d", &t); 40 for (int caseNum = 1; caseNum <= t; caseNum++) { 41 printf("Case #%d: ", caseNum); 42 for (int i = 0; i < 10; i++) a[i] = i; 43 int n; scanf("%d", &n); 44 for (int i = 1; i <= n; i++) cin >> s[i]; 45 sot(s, n); 46 n = unique(s + 1, s + 1 + n) - s - 1; 47 int tmp = min(n, 20), flag = 1; 48 do { 49 int haveSolution = 1; 50 for (int i = 1; haveSolution && i <= tmp; i++) 51 if (!check(i)) haveSolution = 0; 52 if (haveSolution) 53 for (int i = tmp + 1; haveSolution && i <= n; i++) 54 if (!check(i)) haveSolution = 0; 55 if (haveSolution) { 56 for (int i = 0; i < 10; i++) printf("%d", a[i]); 57 puts(""); 58 flag = 0; 59 break; 60 } 61 } while (next_permutation(a, a + 10)); 62 if (flag) puts("Impossible"); 63 } 64 return 0; 65 }
J：
一个人有n个技能，每个技能最高可以升到m级。初始状态下n个技能的等级都是0级。第i个技能从(j-1)级升到j级要花费cij点代价(有可能是负数)。如果所有技能都达到了j级，就可以获得dj点代价(有可能是负数)。输出这个人能获得的最大代价是多少。
这个题很容易想到贪心，然后就踩坑了：枚举有j个等级升满，然后对第i种技能从pre[i][j]到pre[i][m]中选择最小的那个作为第i种等级的最终等级。pre[i][j]表示第i种技能升j级所需代价和。这种做法的问题在于d[j]可能是负的，贪心选最小的pre可能导致j+1之后某些负的level使得答案变小。可以这样修正得到正解：枚举一种等级i作为level最低的技能，并枚举它的等级j，那么其他技能的等级可以在j到m之间任取。
1 /* basic header */ 2 #include <bits/stdc++.h> 3 /* define */ 4 #define ll long long 5 #define dou double 6 #define pb emplace_back 7 #define mp make_pair 8 #define sot(a,b) sort(a+1,a+1+b) 9 #define rep1(i,a,b) for(int i=a;i<=b;++i) 10 #define rep0(i,a,b) for(int i=a;i<b;++i) 11 #define eps 1e-8 12 #define int_inf 0x3f3f3f3f 13 #define ll_inf 0x7f7f7f7f7f7f7f7f 14 #define lson (curpos<<1) 15 #define rson (curpos<<1|1) 16 /* namespace */ 17 using namespace std; 18 /* header end */ 19 20 const int maxn = 1e3 + 10; 21 ll c[maxn][maxn], d[maxn], sum[maxn], val[maxn], tem[maxn]; 22 int n, m; 23 24 int pos(ll *c, int l, int r) { 25 ll val = ll_inf; int minI; 26 for (int i = l; i <= r; i++) 27 if (c[i] <= val) { 28 val = c[i]; 29 minI = i; 30 } 31 return minI; 32 } 33 34 int main() { 35 int t; scanf("%d", &t); 36 for (int casenum = 1; casenum <= t; casenum++) { 37 scanf("%d%d", &n, &m); 38 for (int i = 1; i <= n; i++) sum[i] = 0; 39 for (int i = 1; i <= n; i++) { 40 c[i][0] = 0; val[i] = ll_inf; 41 for (int j = 1; j <= m; j++) { 42 scanf("%lld", &c[i][j]); 43 sum[j] += c[i][j]; 44 c[i][j] += c[i][j - 1]; 45 if (c[i][j] <= val[i]) { 46 val[i] = c[i][j]; 47 tem[i] = j; 48 } 49 } 50 } 51 d[0] = 0; 52 for (int i = 1; i <= m; i++) { 53 scanf("%lld", &d[i]); 54 d[i] -= sum[i]; d[i] += d[i - 1]; 55 } 56 ll ans = 0, tmp, maxv; 57 for (int i = 0; i <= m; i++) { 58 tmp = d[i]; int cnt = 0; 59 if (i != m) { 60 ll minv; maxv = -ll_inf; 61 for (int j = 1; j <= n; j++) { 62 if (tem[j] <= i) tem[j] = pos(c[j], i + 1, m); 63 minv = c[j][tem[j]]; 64 if (minv - c[j][i] < 0) { 65 maxv = max(maxv, minv - c[j][i]); 66 tmp -= minv - c[j][i]; 67 cnt++; 68 } 69 } 70 } 71 if (cnt == n) tmp += maxv; 72 ans = max(ans, tmp); 73 } 74 printf("Case #%d: %lld ", casenum, ans); 75 } 76 return 0; 77 }
也可以这样做：显然升级过程没有后效性，考虑dp。设dp[i][j]为前i种科技最小等级为j时，所付出的最小代价。
状态转移方程dp[i+1][j]=min(dp[i][k]+pre[i+1][u])，其中k>=j，u>=j，且k和u至少有一个为j。
1 /* basic header */ 2 #include <bits/stdc++.h> 3 /* define */ 4 #define ll long long 5 #define dou double 6 #define pb emplace_back 7 #define mp make_pair 8 #define sot(a,b) sort(a+1,a+1+b) 9 #define rep1(i,a,b) for(int i=a;i<=b;++i) 10 #define rep0(i,a,b) for(int i=a;i<b;++i) 11 #define eps 1e-8 12 #define int_inf 0x3f3f3f3f 13 #define ll_inf 0x7f7f7f7f7f7f7f7f 14 #define lson (curpos<<1) 15 #define rson (curpos<<1|1) 16 /* namespace */ 17 using namespace std; 18 /* header end */ 19 20 const int maxn = 2e3 + 10; 21 ll a[maxn][maxn], sum[maxn], dp[maxn][maxn][20]; 22 int n, m; 23 24 void init() { 25 for (int i = 1; i <= n; i++) 26 for (int j = 1; j <= m + 1; j++) 27 dp[i][j][0] = a[i][j]; 28 for (int k = 1; k <= n; k++) 29 for (int j = 1; (1 << j) <= m + 1; j++) 30 for (int i = 1; i + (1 << j) - 1 <= m + 1; i++) 31 dp[i][k][j] = min(dp[i][k][j - 1], dp[i][k + (1 << j - 1)][j - 1]); 32 } 33 34 ll rmq(int j, int l, int r) { 35 int k = log2(r - l + 1); 36 return min(dp[j][l][k], dp[j][r - (1 << k) + 1][k]); 37 } 38 39 int main() { 40 int t; scanf("%d", &t); 41 for (int caseNum = 1; caseNum <= t; caseNum++) { 42 scanf("%d%d", &n, &m); 43 for (int i = 1; i <= n; i++) 44 for (int j = 2; j <= m + 1; j++) { 45 scanf("%lld", &a[i][j]); // 同一技能升级代价做前缀和 46 a[i][j] += a[i][j - 1]; 47 } 48 init(); 49 for (int i = 2; i <= m + 1; i++) { 50 scanf("%lld", &sum[i]); // 技能最低等级奖励也做前缀和 51 sum[i] += sum[i - 1]; 52 } 53 ll ans = 0; 54 for (int j = 1; j <= m + 1; j++) { 55 ll tmp1 = 0; 56 for (int i = 1; i <= n; i++) tmp1 += rmq(i, j, m + 1); 57 for (int i = 1; i <= n; i++) { 58 ll tmp = tmp1; 59 tmp -= rmq(i, j, m + 1); tmp += a[i][j]; 60 tmp = sum[j] - tmp; ans = max(ans, tmp); 61 } 62 } 63 printf("Case #%d: %lld ", caseNum, ans); 64 } 65 return 0; 66 }