Codeforces Round #419 (Div. 2)

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A题

题意:
给定一个时间，求下一个特定时间之前要睡的时间(回文串)

顺序模拟下就可

#include <bits/stdc++.h>using namespace std;char s[6];int d[4];void tran(char *s){    if(++s[3] > '9'){        s[3] = '0';        if(++s[2] > '5'){            s[2] = '0';            ++s[1];            if(s[0] == '2' && s[1] == '4'){                s[0] = s[1] = '0';            }            if(s[1] > '9'){                s[1] = '0';                ++s[0];            }        }    }}bool test(char *s){    return s[0] == s[3] && s[1] == s[2];}int main(){   // freopen("ce.in", "r", stdin);    cin >> s;    s[2] = s[3];    s[3] = s[4];    s[4] = 0;    int ans = 0;    while(!test(s)) ++ans, tran(s);    cout << ans << endl;    return 0;}

B题

[x, y]染色，差分数组:
++d[x], –d[y + 1];
d[i] = a[i] - [i - 1];
a[0] = 0;

所以a[i]=∑ii=1d[i]
前缀和:
令sum[i]为[1, n]中满足的个数
则sum[i]=∑ni=1(a[i]>=k)?1:0

则对于查询[x, y],
ans = sump[y] - sum[x - 1]

当时用线段树做的..

#include <bits/stdc++.h>using namespace std;const int N = 200010;int a[200010];int main(){    //freopen("ce.in", "r", stdin);    memset(a, 0, sizeof a);    int n, k, q, l, r;    cin >> n >> k >> q;    while(n--){        cin >> l >> r;        ++a[l], --a[r + 1];    }    for(int i = 1; i < 200001; ++i) a[i] += a[i - 1];    for(int i = 1; i < 200001; ++i) {            a[i] = a[i] >= k ? 1:0;            a[i] += a[i - 1];    }    while(q--){        cin >> l >> r;        cout << a[r] - a[l - 1] << endl;    }    return 0;}

C题

1首先比较行数, 列数；
较大的优先删;

2存每一行/列的multiset
3取第一行/列的最大最小，如果两值相等, 则判断每一行/列是不是都相同,
如果相同按数目删行/列, 不同无解；
第一行/列两值不相等, 找最大的书局那列/行删, 记下列/行号；
重复3直至两值相等

code：

#include <bits/stdc++.h>using namespace std;int n;int g[110][110];int ansr[110];multiset<int> s[110];map<int, set<int> >mp;multiset<int> ansl;int main(){    //freopen("ce.in", "r", stdin);    int r, c;    scanf("%d%d", &r, &c);    for(int i = 1; i <= r; ++i)        for(int j = 1; j <= c; ++j)        {            scanf("%d", &g[i][j]);            //   printf("%d\n", g[i][j]);        }    if(r >= c)    {        for(int i = 1; i <= c; ++i)            for(int j = 1; j <= r; ++j)            {                s[i].insert(g[j][i]);            }        for(int i = 1; i <= r; ++i) mp[g[i][1]].insert(i);        multiset<int>::iterator op, ed;        set<int>::iterator it;        op = s[1].begin();        ed = --s[1].end();        int sum = 0;        while(*op < *ed)        {            //     printf("%d %d\n", *op, *ed);            int v = *ed;            it = --mp[v].end();            int row = *it;            ansl.insert(row);            for(int i = 1; i <= c; ++i)            {                s[i].erase(--s[i].end());                s[i].insert(g[row][i] - 1);                if(--g[row][i] < 0)                {                    printf("-1\n");                    return 0;                }            }            mp[v].erase(row);            mp[v - 1].insert(row);            ++sum;            op = s[1].begin();            ed = --s[1].end();            //  printf("%d %d\n", *op, *ed);        }        for(int i = 1; i <= c; ++i)        {            op = s[i].begin();            ed = --s[i].end();            if(*op < *ed)            {                printf("-1\n");                return 0;            }            ansr[i] = *op;            sum += ansr[i];        }        printf("%d\n", sum);        for(op = ansl.begin(); op != ansl.end(); ++op)            printf("row %d\n", *op);        for(int i = 1; i <= c; ++i)        {            for(int j = 1; j <= ansr[i]; ++j)            {                printf("col %d\n", i);            }        }    }else{        for(int i = 1; i <= r; ++i)            for(int j = 1; j <= c; ++j)            {                s[i].insert(g[i][j]);            }        for(int i = 1; i <= c; ++i) mp[g[1][i]].insert(i);        multiset<int>::iterator op, ed;        set<int>::iterator it;        op = s[1].begin();        ed = --s[1].end();        int sum = 0;        while(*op < *ed)        {            //     printf("%d %d\n", *op, *ed);            int v = *ed;            it = --mp[v].end();            int col = *it;            ansl.insert(col);            for(int i = 1; i <= r; ++i)            {                s[i].erase(--s[i].end());                s[i].insert(g[i][col] - 1);                if(--g[i][col] < 0)                {                    printf("-1\n");                    return 0;                }            }            mp[v].erase(col);            mp[v - 1].insert(col);            ++sum;            op = s[1].begin();            ed = --s[1].end();            //  printf("%d %d\n", *op, *ed);        }        for(int i = 1; i <= r; ++i)        {            op = s[i].begin();            ed = --s[i].end();            if(*op < *ed)            {                printf("-1\n");                return 0;            }            ansr[i] = *op;            sum += ansr[i];        }        printf("%d\n", sum);        for(op = ansl.begin(); op != ansl.end(); ++op)            printf("col %d\n", *op);        for(int i = 1; i <= r; ++i)        {            for(int j = 1; j <= ansr[i]; ++j)            {                printf("row %d\n", i);            }        }    }    return 0;}

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