Codeforces Round #312 (Div. 2) 完整题解

A题：sb题。。。。就判断一下哪边最大即可。。

#include <iostream>#include <queue>#include <stack>#include <map>#include <set>#include <bitset>#include <cstdio>#include <algorithm>#include <cstring>#include <climits>#include <cstdlib>#include <cmath>#include <time.h>#define maxn 305#define maxm 20005#define eps 1e-7#define mod 1000000007#define INF 0x3f3f3f3f#define PI (acos(-1.0))#define lowbit(x) (x&(-x))#define mp make_pair#define ls o<<1#define rs o<<1 | 1#define lson o<<1, L, mid #define rson o<<1 | 1, mid+1, R#define pii pair<int, int>#pragma comment(linker, "/STACK:16777216")typedef long long LL;typedef unsigned long long ULL;//typedef int LL; using namespace std;LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}// headpii p[maxn];int n;void work(){int x, a;scanf("%d", &n);for(int i = 1; i <= n; i++) {scanf("%d%d", &x, &a);p[i] = mp(x, a);}sort(p+1, p+n+1);int t = lower_bound(p+1, p+n+1, mp(0, 0)) - p - 1;int t2 = n - t;t2 = min(t, t2);int ans1 = 0, ans2 = 0;for(int i = t; i >= max(1, t - t2); i--) {ans1 += p[i].second;}for(int i = t+1; i <= min(n, t + t2); i++) {ans1 += p[i].second;}for(int i = t; i >= max(1, t - t2 + 1); i--) {ans2 += p[i].second;}for(int i = t+1; i <= min(n, t + t2 + 1); i++) {ans2 += p[i].second;}printf("%d\n", max(ans1, ans2));}int main(){work();return 0;}

B题：sb题。。判断一下每个数的最左最右。。

#include <iostream>#include <queue>#include <stack>#include <map>#include <set>#include <bitset>#include <cstdio>#include <algorithm>#include <cstring>#include <climits>#include <cstdlib>#include <cmath>#include <time.h>#define maxn 1000005#define maxm 2000005#define eps 1e-7#define mod 1000000007#define INF 0x3f3f3f3f#define PI (acos(-1.0))#define lowbit(x) (x&(-x))#define mp make_pair#define ls o<<1#define rs o<<1 | 1#define lson o<<1, L, mid #define rson o<<1 | 1, mid+1, R#define pii pair<int, int>#pragma comment(linker, "/STACK:16777216")typedef long long LL;typedef unsigned long long ULL;//typedef int LL; using namespace std;LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}// headint a[maxn];int c1[maxn];int c2[maxn];int cnt[maxn];int n;void work(){scanf("%d", &n);for(int i = 1; i <= n; i++) {scanf("%d", &a[i]);}memset(c1, 0, sizeof c1);memset(c2, 0, sizeof c2);for(int i = 1; i <= n; i++) c2[a[i]] = i;for(int i = n; i >= 1; i--) c1[a[i]] = i;for(int i = 1; i <= n; i++) cnt[a[i]]++;int mx = 0;for(int i = 1; i <= 1e6; i++) mx = max(mx, cnt[i]);int ans1 = -1, ans2 = -1, ans = INF;for(int i = 1; i <= n; i++) {if(cnt[a[i]] != mx) continue;int t = c2[a[i]] - c1[a[i]] + 1;if(t < ans) {ans = t;ans1 = c1[a[i]];ans2 = c2[a[i]];}}printf("%d %d\n", ans1, ans2);}int main(){work();return 0;}

C题：有点思维难度。。。先搞出可以到达的最终数字的最小值。。。然后处理出每个数到这个最小值只剩2的倍数的最小值。。。再处理这个数还要除掉几次到达最小值。。取中位数就行了。。

#include <iostream>#include <queue>#include <stack>#include <map>#include <set>#include <bitset>#include <cstdio>#include <algorithm>#include <cstring>#include <climits>#include <cstdlib>#include <cmath>#include <time.h>#define maxn 100005#define maxm 200005#define eps 1e-7#define mod 1000000007#define INF 0x3f3f3f3f#define PI (acos(-1.0))#define lowbit(x) (x&(-x))#define mp make_pair#define ls o<<1#define rs o<<1 | 1#define lson o<<1, L, mid #define rson o<<1 | 1, mid+1, R#define pii pair<int, int>#pragma comment(linker, "/STACK:16777216")typedef long long LL;typedef unsigned long long ULL;//typedef int LL; using namespace std;LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}// headmultiset<int> s;multiset<int>::iterator it;map<int, int> mpp;int a[maxn];int n;void work(){scanf("%d", &n);for(int i = 1; i <= n; i++) scanf("%d", &a[i]);for(int i = 1; i <= n; i++) {mpp[a[i]]++;s.insert(a[i]);}int t;while(1) {t = *(--s.end());if(mpp[t] == n) break;mpp[t]--;it = s.find(t);s.erase(it);mpp[t / 2]++;s.insert(t / 2);}int ans = 0;for(int i = 1; i <= n; i++) {int tt = t;while(tt <= a[i]) tt <<= 1;tt /= 2;tt ^= a[i];int res = 0;while(tt) res++, tt >>= 1;ans += res;for(int j = 0; j < res; j++) a[i] /= 2;res = 0;while(a[i] != t) a[i] /= 2, res++;a[i] = res;}sort(a+1, a+n+1);int tt = a[(n + 1) / 2];for(int i = 1; i <= n; i++) ans += abs(a[i] - tt);printf("%d\n", ans);}int main(){work();return 0;}

D题：把真拆成左右两个假。。。然后排序，判断一下就行了。

#include <iostream>#include <queue>#include <stack>#include <map>#include <set>#include <bitset>#include <cstdio>#include <algorithm>#include <cstring>#include <climits>#include <cstdlib>#include <cmath>#include <time.h>#define maxn 400005#define maxm 400005#define eps 1e-7#define mod 1000000007#define INF 0x3f3f3f3f#define PI (acos(-1.0))#define lowbit(x) (x&(-x))#define mp make_pair#define ls o<<1#define rs o<<1 | 1#define lson o<<1, L, mid #define rson o<<1 | 1, mid+1, R#define pii pair<int, int>#pragma comment(linker, "/STACK:16777216")typedef long long LL;typedef unsigned long long ULL;//typedef int LL; using namespace std;LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}// headpair<LL, LL> a[maxn];LL two[maxn];LL sum[maxn];int n, h, cnt;void work(){int id, op;LL L, R;scanf("%d%d", &h, &n);cnt = 0;two[1] = 1;for(int i = 2; i <= h; i++) two[i] = two[i-1] * 2;for(int i = 1; i <= h; i++) sum[i] = sum[i-1] + two[i];for(int i = 1; i <= n; i++) {scanf("%d%I64d%I64d%d", &id, &L, &R, &op);L -= sum[id-1];R -= sum[id-1];while(id < h) {id++;L = 2 * L - 1;R = 2 * R;}if(op == 0) a[cnt++] = mp(L, R);else {if(L > 1) a[cnt++] = mp(1, L-1);if(R < two[h]) a[cnt++] = mp(R+1, two[h]);}}LL ans = -1, now = 1, t = 0;sort(a, a+cnt);for(int i = 0; i < cnt; i++) {if(a[i].first > now) {ans = now;t += a[i].first - now;}now = max(now, a[i].second + 1);}if(two[h] >= now) {ans = now;t += two[h] - now + 1;}if(t == 0) printf("Game cheated!\n");else if(t > 1) printf("Data not sufficient!\n");else printf("%I64d\n", ans + sum[h-1]);}int main(){work();return 0;}

E题：线段树。。先按字母建立26棵线段树。。。每次操作就是对该区间的26棵线段树先26次区间更新，再26次按升序或降序区间更新。。。

#include <iostream>#include <queue>#include <stack>#include <map>#include <set>#include <bitset>#include <cstdio>#include <algorithm>#include <cstring>#include <climits>#include <cstdlib>#include <cmath>#include <time.h>#define maxn 100005#define maxm 200005#define eps 1e-7#define mod 1000000007#define INF 0x3f3f3f3f#define PI (acos(-1.0))#define lowbit(x) (x&(-x))#define mp make_pair#define ls o<<1#define rs o<<1 | 1#define lson o<<1, L, mid #define rson o<<1 | 1, mid+1, R#define pii pair<int, int>#pragma comment(linker, "/STACK:16777216")typedef long long LL;typedef unsigned long long ULL;//typedef int LL; using namespace std;LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}// headchar s[maxn];int Sum[26][maxn << 2];int Lazy[26][maxn << 2];int a[maxn];int c[30];int n, m;void pushup(int sum[], int o){sum[o] = sum[ls] + sum[rs];}void pushdown(int sum[], int lazy[], int o, int L, int R){if(lazy[o] != -1) {int mid = (L + R) >> 1;sum[ls] = (mid - L + 1) * lazy[o];sum[rs] = (R - mid) * lazy[o];lazy[ls] = lazy[rs] = lazy[o];lazy[o] = -1;}}void build(int sum[], int lazy[], int o, int L, int R, int rt){lazy[o] = -1;if(L == R) {if(rt == a[L]) sum[o] = 1;else sum[o] = 0;return;}int mid = (L + R) >> 1;build(sum, lazy, lson, rt);build(sum, lazy, rson, rt);pushup(sum, o);}void update(int sum[], int lazy[], int o, int L, int R, int ql, int qr, int k){if(ql <= L && qr >= R) {sum[o] = k * (R - L + 1);lazy[o] = k;return;}pushdown(sum, lazy, o, L, R);int mid = (L + R) >> 1;if(ql <= mid) update(sum, lazy, lson, ql, qr, k);if(qr > mid) update(sum, lazy, rson, ql, qr, k);pushup(sum, o);}int query(int sum[], int lazy[], int o, int L, int R, int ql, int qr){if(ql <= L && qr >= R) return sum[o];pushdown(sum, lazy, o, L, R);int mid = (L + R) >> 1, ans = 0;if(ql <= mid) ans += query(sum, lazy, lson, ql, qr);if(qr > mid) ans += query(sum, lazy, rson, ql, qr);pushup(sum, o);return ans;}void solve(int l, int r, int op){memset(c, 0, sizeof c);for(int i = 0; i < 26; i++) {c[i] += query(Sum[i], Lazy[i], 1, 1, n, l, r);update(Sum[i], Lazy[i], 1, 1, n, l, r, 0);}int t = l;if(op == 0) {for(int i = 25; i >= 0; i--) {if(c[i] == 0) continue;update(Sum[i], Lazy[i], 1, 1, n, t, t + c[i] - 1, 1);t += c[i];}}else {for(int i = 0; i < 26; i++) {if(c[i] == 0) continue;update(Sum[i], Lazy[i], 1, 1, n, t, t + c[i] - 1, 1);t += c[i];}}}void work(){scanf("%s", s + 1);for(int i = 1; i <= n; i++) a[i] = s[i] - 'a';for(int i = 0; i < 26; i++) build(Sum[i], Lazy[i], 1, 1, n, i);for(int i = 1; i <= m; i++) {int l, r, kk;scanf("%d%d%d", &l, &r, &kk);solve(l, r, kk);}for(int i = 1; i <= n; i++) {for(int j = 0; j < 26; j++) {int t = query(Sum[j], Lazy[j], 1, 1, n, i, i);if(t) {s[i] = j + 'a';break;}}}s[n+1] = '\0';puts(s + 1);}int main(){while(scanf("%d%d", &n, &m) != EOF) {work();}return 0;}