青岛理工大学第五届ACM交流赛 部分题解

A：后缀维护si*pi的最小值，查询的时候二分，判断后缀和当前两个部分就行。

#include <bits/stdc++.h>
using namespace std;
 
typedef long long LL;
const int maxn = 100100;
int n, m;
LL s[maxn], p[maxn];
LL suf[maxn];
 
int main() {
    // freopen("in", "r", stdin);
    int T;
    LL q;
    scanf("%d", &T);
    while(T--) {
        scanf("%d%d",&n,&m);
        memset(suf, 0, sizeof(suf));
        for(int i = 0; i < n; i++) scanf("%lld%lld",&s[i],&p[i]);
        suf[n-1] = s[n-1] * p[n-1];
        for(int i = n - 2; i >= 0; i--) suf[i] = min(suf[i+1], s[i]*p[i]);
        s[n] = 100000000000000LL; suf[n] = 100000000000000LL;
        while(m--) {
            scanf("%lld", &q);
            int pos = lower_bound(s, s+n+1, q) - s;
            // if(s[pos] == q) printf("%lld\n", q*p[pos]);
            if(pos == n) printf("%lld\n", q*p[n-1]);
            else printf("%lld\n", min(suf[pos], q*p[pos-1]));
        }
    }
    return 0;
}

 

B：先离散化找出所有内容的种类，之后尺取卡出最小页数。

#include <bits/stdc++.h>
using namespace std;
 
const int maxn = 1001000;
int n, m, rl, rr;
int h[maxn], p[maxn];
int vis[maxn];
int cnt, ret;
 
int id(int x) {
    return lower_bound(h, h+m, x) - h;
}
 
int main() {
    // freopen("in", "r", stdin);
    while(~scanf("%d", &n)) {
        for(int i = 0; i < n; i++) {
            scanf("%d", &p[i]);
            h[i] = p[i];
        }
        sort(h,  h+n); m = unique(h, h+n) - h;
        memset(vis, 0, sizeof(vis));
        cnt = 0; ret = n; rl = 0, rr = n - 1;
        int lo = 0, hi = 0;
        while(lo < n) {
            while(hi < n && cnt < m) {
                int idx = id(p[hi++]);
                if(!vis[idx]) cnt++;
                vis[idx]++;
            }
            if(cnt < m) break;
            if(ret > hi - lo && cnt == m) {
                ret = hi - lo;
                rl = lo, rr = hi - 1;
            }
            // printf("%d %d\n", lo, hi);
            int idx = id(p[lo++]);
            vis[idx]--;
            if(vis[idx] == 0) cnt--;
            if(lo >= n) break;
        }
        printf("%d\n", ret);
        printf("%d %d\n", rl, rr);
    }
    return 0;
}

 

C：水题…

#include <bits/stdc++.h>
using namespace std;

const int maxn = 100100;
int n;
int a[maxn], b[maxn];

int main() {
    // freopen("in", "r", stdin);
    int T;
    scanf("%d", &T);
    while(T--) {
        scanf("%d", &n);
        int x = 0, y = 0;
        for(int i = 1; i <= n; i++) {
            scanf("%d%d",&a[i],&b[i]);
        }
        sort(a+1, a+n+1);
        sort(b+1, b+n+1);
        for(int i = 1; i <= n; i++) {
            if(a[i] > b[i]) x += 2;
            else if(a[i] == b[i]) x++, y++;
            else y += 2;
        }
        if(x > y) puts("Winjourn");
        else if(x == y) puts("Tie");
        else puts("Never_Sorry");
    }
    return 0;
}

 

D：数位DP，dp(l,cnt)统计长度为l到时候1的个数为cnt的整数。

#include <bits/stdc++.h>
using namespace std;
 
typedef long long LL;
const int maxn = 22;
int digit[maxn];
LL dp[maxn][maxn];
LL n;
 
LL dfs(int l, int cnt, bool flag) {
    if(l == 0) return cnt;
    if(!flag && ~dp[l][cnt]) return dp[l][cnt];
    int pos = flag ? digit[l] : 9;
    LL ret = 0;
    for(int i = 0; i <= pos; i++) {
        ret += dfs(l-1, cnt+(i==1), flag&&(pos==i));
    }
    if(!flag) dp[l][cnt] = ret;
    return ret;
}
 
LL f(LL x) {
    int pos = 0;
    while(x) {
        digit[++pos] = x % 10;
        x /= 10;
    }
    return dfs(pos, 0, true);
}
 
int main() {
    // freopen("in", "r", stdin);
    int _ = 1;
    memset(dp, -1, sizeof(dp));
    while(~scanf("%lld", &n)) {
        printf("Case %d:%lld\n", _++, f(n));
    }
    return 0;
}

 

E：模拟，如果循环节正好长度为50的话，就不要打省略号了。。

#include <bits/stdc++.h>
using namespace std;
 
vector<int> t1, t2;
map<int, int> id;
int a, b;
 
void f(int a, int b) {
    id.clear();
    t1.clear(); t2.clear();
    int p = -1;
    while(1) {
        t1.push_back(a/b);
        int x = a % b;
        if(x == 0) break;
        if(id.find(x) == id.end()) {
            id[x] = t2.size();
            t2.push_back(x);
        }
        else {
            p = id[x];
            break;
        }
        a = x * 10;
    }
    if(t1.size() == 0) while(1);
    printf("%d.", t1[0]);
    bool flag = 0;
    bool plusflag = 0;
    int cnt = 0;
    for(int i = 1; i < t1.size(); i++) {
        if(cnt >= 50) {
            plusflag = 1;
            break;
        }
        if(i == p + 1) {
            printf("(");
            flag = 1;
        }
        printf("%d", t1[i]);
        if(flag) cnt++;
    }
    if(plusflag) printf("...");
    if(p >= 0) {
        if(flag) printf(")\n");
        printf("%d\n", t1.size()-p-1);
    }
    else {
        printf("(0)\n");
        printf("1\n");
    }
}
 
int main() {
    // freopen("in", "r", stdin);
    // freopen("out", "w", stdout);
    int _ = 1;
    while(~scanf("%d%d",&a,&b)) {
        printf("Case %d:\n%d/%d=", _++, a, b);
        if(a % b == 0) printf("%d.(0)\n1\n", a / b);
        else f(a, b);
    }
    return 0;
}

 

F：好麻烦的计算几何，感觉要先卡出封闭图形，之后找到所有的矩形。然后看看能放的最大圆。

G：对整个数组排序后二分答案，贪心地从左到右选取数字，保证相邻两个数的差不小于二分出来的值，判断是否满足大于等于k个。

#include <bits/stdc++.h>
using namespace std;
 
const int maxn = 100100;
int n, k;
int a[maxn], s[maxn];
int ret;
 
bool ok(int dis) {
    int i = 2, p = 1;
    int cnt = 0;
    while(i <= n) {
        if(a[i] - a[p] >= dis) {
            p = i; i++;
            cnt++;
        }
        else i++;
    }
    return cnt + 1 >= k;
}
 
int main() {
    // freopen("in", "r", stdin);
    while(~scanf("%d%d",&n,&k)) {
        memset(s, 0, sizeof(s));
        for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
        sort(a+1, a+n+1);
        ret = a[n] - a[1];
        int lo = 0, hi = a[n];
        while(lo <= hi) {
            int mid = (lo + hi) >> 1;
            if(ok(mid)) {
                ret = mid;
                lo = mid + 1;
            }
            else hi = mid - 1;
        }
        printf("%d\n", ret);
    }
    return 0;
}

 

H：像是需要用启发式搜索将整座附近的点处理出来一个优先值，顺着第一条路走，最后回来。（后来有巨巨说是BFS）

I：枚举a,b字符串子串的起始，扫描对应位，如果不相同则替换至相同，最多k次操作。贪心地将替换操作放到前面，如果不能替换则break更新ret。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <map>
using namespace std;

const int maxn = 550;
int n, m, k;
char a[maxn], b[maxn];

int main() {
    // freopen("in" ,"r", stdin);
    while(~scanf("%d", &k)) {
        scanf("%s%s",a,b);
        n = strlen(a); m = strlen(b);
        int ret = 0;
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < m; j++) {
                int cnt = 0, tmp = 0;
                for(int aa = i, bb = j; aa < n && bb < m; aa++, bb++) {
                    if(a[aa] == b[bb]) tmp++;
                    else {
                        if(cnt < k) tmp++, cnt++;
                        else break;
                    }
                }
                ret = max(ret, tmp);
            }
        }
        printf("%d\n", ret);
    }
    return 0;
}

 

J：f(i,j)表示和为i，用j个数字组成的方案数，转移方程f(i,j)=f(i-p,j-1)，p取1~4，会爆LL所以用double存了f。

#include <bits/stdc++.h>
using namespace std;
 
typedef long long LL;
const int maxn = 111;
const int maxm = 666;
int n, m;
double f[maxn][maxm];
 
int main() {
    // freopen("in", "r", stdin);
    memset(f, 0, sizeof(f));
    f[1][1] = 1.0; f[1][2] = 1.0; f[1][3] = 1.0; f[1][4] = 1.0;
    for(int i = 2; i < maxn; i++) {
        for(int j = 1; j < maxm; j++) {
            for(int k = 1; k <= 4; k++) {
                if(k >= j) continue;
                f[i][j] += f[i-1][j-k];
            }
        }
    }
    while(~scanf("%d%d",&n,&m)) {
        printf("%.4lf\n", f[n][m]/pow(4,n));
    }
    return 0;
}

 

K：最后的情形无非是两种，一种是wywy..，另一种是ywyw..。整个串长度如果是奇数那么是无法围成要求的项链的；之后判断当前情况和上述哪种情况最近，然后分别找w和y错位的数量，先交换，再修改，min(w,y)+(max(w,y)-min(w,y))=max(w,y)即为答案。

#include <bits/stdc++.h>
using namespace std;

const int maxn = 100100;
char s[maxn], t[2][maxn];
int n;

int main() {
    // freopen("in", "r", stdin);
    memset(t, 0, sizeof(t));
    for(int i = 0; i < maxn; i++) {
        if(i % 2 == 0) {
            t[0][i] = 'w', t[1][i] = 'y';
        }
        else {
            t[0][i] = 'y', t[1][i] = 'w';
        }
    }
    while(~scanf("%d", &n)) {
        scanf("%s", s);
        if(n & 1) {
            puts("What a pity!");
            continue;
        }
        int a = 0, b = 0;
        for(int i = 0; i < n; i++) {
            if(s[i] != t[0][i]) a++;
            if(s[i] != t[1][i]) b++;
        }
        if(a < b) {
            int w = 0, y = 0;
            for(int i = 0; i < n; i++) {
                if(s[i] != t[0][i]) {
                    if(s[i] == 'w') w++;
                    else y++;
                }
            }
            printf("%d\n", max(w, y));
        }
        else {
            int w = 0, y = 0;
            for(int i = 0; i < n; i++) {
                if(s[i] != t[1][i]) {
                    if(s[i] == 'w') w++;
                    else y++;
                }
            }
            printf("%d\n", max(w, y));
        }
    }
    return 0;
}

 

L：不会…