考试只出一道题。。。
A>
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #define rep(i, a, b) for (int i = a; i <= b; i++) 5 #define drep(i, a, b) for (int i = a; i >= b; i--) 6 #define REP(i, a, b) for (int i = a; i < b; i++) 7 #define mp make_pair 8 #define pb push_back 9 #define clr(x) memset(x, 0, sizeof(x)) 10 #define xx first 11 #define yy second 12 13 using namespace std; 14 15 typedef long long i64; 16 typedef pair<int, int> pii; 17 const int inf = ~0U >> 1; 18 const i64 INF = ~0ULL >> 1; 19 //***************************** 20 21 int a, b, c; 22 int main() { 23 scanf("%d%d%d", &a, &b, &c); 24 int ans = inf; 25 ans = min(ans, (a << 1) + (b << 1)); 26 ans = min(ans, (a << 1) + (c << 1)); 27 ans = min(ans, (b << 1) + (c << 1)); 28 ans = min(ans, a + b + c); 29 printf("%d ", ans); 30 return 0; 31 }
B> 注意不是一一对应
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #define rep(i, a, b) for (int i = a; i <= b; i++) 5 #define drep(i, a, b) for (int i = a; i >= b; i--) 6 #define REP(i, a, b) for (int i = a; i < b; i++) 7 #define mp make_pair 8 #define clr(x) memset(x, 0, sizeof(x)) 9 #define xx first 10 #define yy second 11 12 using namespace std; 13 14 typedef long long i64; 15 typedef pair<int, int> pii; 16 const int inf = ~0U >> 1; 17 const i64 INF = ~0ULL >> 1; 18 //***************************** 19 20 const int maxn = 100005; 21 int a[maxn], b[maxn], ta[maxn], tb[maxn]; 22 pii pa[maxn], pb[maxn]; 23 int main() { 24 int n, m; 25 scanf("%d%d", &n, &m); 26 rep(i, 1, n) scanf("%d", &a[i]), ta[a[i]]++, pa[i] = mp(a[i], i); 27 rep(i, 1, m) scanf("%d", &b[i]), tb[b[i]]++, pb[i] = mp(b[i], i); 28 rep(i, 1, n) if (!ta[i] && tb[i]) { 29 puts("Impossible"); 30 return 0; 31 } 32 rep(i, 1, m) if (ta[b[i]] > 1) { 33 puts("Ambiguity"); 34 return 0; 35 } 36 sort(pa + 1, pa + 1 + n), sort(pb + 1, pb + 1 + m); 37 static int ans[maxn]; 38 rep(i, 1, m) { 39 int pos = lower_bound(pa + 1, pa + 1 + n, mp(pb[i].xx, -1)) - pa; 40 ans[pb[i].yy] = pa[pos].yy; 41 } 42 puts("Possible"); 43 rep(i, 1, m) printf("%d ", ans[i]); 44 return 0; 45 }
C>
树状数组维护比一个点小的值最远在哪里。
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #define rep(i, a, b) for (int i = a; i <= b; i++) 5 #define drep(i, a, b) for (int i = a; i >= b; i--) 6 #define REP(i, a, b) for (int i = a; i < b; i++) 7 #define mp make_pair 8 #define pb push_back 9 #define clr(x) memset(x, 0, sizeof(x)) 10 #define xx first 11 #define yy second 12 13 using namespace std; 14 15 typedef long long i64; 16 typedef pair<int, int> pii; 17 const int inf = ~0U >> 1; 18 const i64 INF = ~0ULL >> 1; 19 //***************************** 20 21 const int maxn = 100005; 22 23 int cas[maxn]; 24 int hsh[maxn], cd; 25 int lst[maxn]; 26 int c[maxn]; 27 void add(int x, int v) { 28 while (x < maxn) { 29 c[x] = max(c[x], v); 30 x += x & (-x); 31 } 32 } 33 int getMax(int x) { 34 int s = 0; 35 while (x > 0) { 36 s = max(s, c[x]); 37 x -= x & (-x); 38 } 39 return s; 40 } 41 int main() { 42 int n; 43 scanf("%d", &n); 44 rep(i, 1, n) scanf("%d", &cas[i]), hsh[cd++] = cas[i]; 45 sort(hsh, hsh + cd); cd = unique(hsh, hsh + cd) - hsh; 46 rep(i, 1, n) cas[i] = lower_bound(hsh, hsh + cd, cas[i]) - hsh + 1; 47 static pii ha[maxn]; 48 rep(i, 1, n) ha[i] = mp(cas[i], i); 49 sort(ha + 1, ha + 1 + n); 50 rep(i, 1, n) { 51 add(ha[i].xx, ha[i].yy); 52 if (ha[i].xx - 1 == 0) lst[ha[i].yy] = i; 53 else lst[ha[i].yy] = getMax(ha[i].xx - 1); 54 lst[ha[i].yy] = max(lst[ha[i].yy], i); 55 } 56 int ans(0); 57 int k = 1; 58 while (k <= n) { 59 ans++; 60 int nx = lst[k]; 61 for (int i = k; i <= nx; i++) if (lst[i] > nx) nx = lst[i]; 62 k = nx + 1; 63 } 64 printf("%d ", ans); 65 }
D>
首先推出公式 n * m 的长方形共有 n * (n + 1) / 2 * (m - n) + n * (n + 1) * (2n + 1) / 6个。
然后发现n * (n + 1) * (2n + 1) 在1442250左右超过了6*1e18，所以有了枚举上界。
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #define rep(i, a, b) for (long long i = a; i <= b; i++) 5 #define drep(i, a, b) for (int i = a; i >= b; i--) 6 #define REP(i, a, b) for (int i = a; i < b; i++) 7 #define mp make_pair 8 #define clr(x) memset(x, 0, sizeof(x)) 9 #define xx first 10 #define yy second 11 12 using namespace std; 13 14 typedef long long i64; 15 typedef pair<int, int> pii; 16 const int inf = ~0U >> 1; 17 const i64 INF = ~0ULL >> 1; 18 //***************************** 19 20 int main() { 21 i64 n; 22 scanf("%I64d", &n); 23 static pair<i64, i64> ans[1442255]; 24 i64 cnt(0); 25 rep(i, 1, 1442250) { 26 i64 tmp = n - i * (i + 1) * (2 * i + 1) / 6; 27 if (tmp % (i * (i + 1) / 2)) continue; 28 if (tmp < 0) continue; 29 i64 m = tmp / (i * (i + 1) / 2) + 1; 30 m = m + i - 1; 31 ans[++cnt] = mp(i, m); 32 if (i != m) ans[++cnt] = mp(m, i); 33 } 34 sort(ans + 1, ans + 1 + cnt); 35 printf("%I64d ", cnt); 36 rep(i, 1, cnt) printf("%I64d %I64d ", ans[i].xx, ans[i].yy); 37 return 0; 38 return 0; 39 }
E>
不会。