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Codeforces Round #422 (Div. 2) A-C

A. I‘m bored with life

水题

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>#include <queue>#include <vector>#include <iomanip>#include <math.h>#include <map>using namespace std;#define FIN     freopen("input.txt","r",stdin);#define FOUT    freopen("output.txt","w",stdout);#define INF     0x3f3f3f3f#define INFLL   0x3f3f3f3f3f3f3f#define lson    l,m,rt<<1#define rson    m+1,r,rt<<1|1typedef long long LL;typedef pair<int, int> PII;using namespace std;int main() {    int a, b;    cin >> a >> b;    int c = min(a, b);    int ans = 1;    for(int i = 1; i <= c; i++) ans *= i;    cout << ans << endl;    return 0;}

 

B. Crossword solving

字符串匹配

英语太差题意花了很久才读懂.....

题意:上面的字符串要把多少个字符变为?才可以变为下面字符串的子串 要变得数量尽可能的小

直接暴力匹配就可以做了

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>#include <queue>#include <vector>#include <iomanip>#include <math.h>#include <map>using namespace std;#define FIN     freopen("input.txt","r",stdin);#define FOUT    freopen("output.txt","w",stdout);#define INF     0x3f3f3f3f#define INFLL   0x3f3f3f3f3f3f3f#define lson    l,m,rt<<1#define rson    m+1,r,rt<<1|1typedef long long LL;typedef pair<int, int> PII;using namespace std;const int maxn = 1000 + 5;char a[maxn];char b[maxn];int p[maxn];int p1[maxn];int main() {    //FIN    int n, m;    scanf("%d%d", &n, &m);    scanf("%s", a);    scanf("%s", b);    int mx = -1;    int c = 0;    for(int i = 0; i <= m - n; i++) {        int num = 0;        int cnt = 0;        for(int j = 0; j < n; j++) {            if(b[i+j] == a[j]) {                num++;            }            else {                p[cnt] = j + 1;                cnt++;            }        }        //cout << num << endl;        if(num > mx) {            mx = num;            for(int j = 0; j < cnt; j++) {                p1[j] = p[j];            }            c = cnt;        }    }    printf("%d\n", n - mx);    for(int i = 0; i < c; i++) printf("%d ", p1[i]);    return 0;}

  

C. Hacker, pack your bags!

结构体一顿瞎做

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>#include <queue>#include <vector>#include <iomanip>#include <math.h>#include <map>using namespace std;#define FIN     freopen("input.txt","r",stdin);#define FOUT    freopen("output.txt","w",stdout);#define INFLL   0x3f3f3f3f3f3f3f#define lson    l,m,rt<<1#define rson    m+1,r,rt<<1|1typedef long long LL;typedef pair<double, double> PII;const long long INF = 1e16;const int maxn = 1e6 + 5;int n, x;LL money[maxn];struct node {    int st, ed, cost, time, flag;}a[maxn];int cmp(node aa, node bb) {    if(aa.st == bb.st) return aa.flag > bb.flag;    else return aa.st < bb.st;}bool check(node aa, node bb) {    if(aa.st < bb.st && aa.ed > bb.st) return 0;    else if(aa.st < bb.st && aa.ed > bb.ed) return 0;    else if(bb.st < aa.st && bb.ed > aa.st) return 0;    else if(bb.st < aa.st && bb.ed > aa.ed) return 0;    else if(bb.st == aa.st || bb.st == aa.ed || bb.ed == aa.st || bb.ed == aa.ed) return 0;    if(aa.time + bb.time != x) return 0;    return 1;}int main() {    //FIN    while(cin >> n >> x) {        int cas = 0;        for(int i = 1; i <= n; i++) {            cin >> a[cas].st >> a[cas].ed >> a[cas].cost;            a[cas].time = a[cas].ed - a[cas].st + 1;            a[cas].flag = 1;            cas++;            a[cas].st = a[cas-1].ed;            a[cas].ed = -1;            a[cas].flag = -1;            a[cas].time = a[cas-1].time;            a[cas].cost = a[cas-1].cost;            cas++;        }        //memset(money, INF, sizeof(money));        for(int i = 0; i <= x; i++) {            money[i] = INF;        }        LL ans = INF;        sort(a, a + cas, cmp);        for(int i = 0; i < cas; i++) {            if(a[i].flag == 1) {                if(x - a[i].time > 0) ans = min(ans , money[x - a[i].time] + (LL)a[i].cost);                //cout <<"i="<<i<<"  "<<money[x-a[i].time]<<endl;                //cout << "ans="<<ans<<endl;            }            else {                    //cout <<"i="<<i<<"  "<<money[x-a[i].time]<<endl;                money[a[i].time] = min(money[a[i].time], (LL)a[i].cost);            }        }        if(ans == INF) cout << "-1" << endl;        else cout << ans << endl;    }    return 0;}

  

 

Codeforces Round #422 (Div. 2) A-C