Codeforces Round #313 (Div. 2)

A
题意：给你n个数，每个数可以使用任意次。问最小的无法得到的正整数，若不存在输出-1。

有1即可。

#include <iostream>#include <cstdlib>#include <cstdio>#include <algorithm>#include <string>#include <map>#include <cstring>#include <stack>#include <queue>#include <cmath>#include <vector>#include <set>using namespace std;typedef long long LL;typedef pair<int, int> pii;int main(){    int n;    while(scanf("%d", &n) != EOF) {        bool flag = false;        for(int i = 1; i <= n; i++) {            int v; scanf("%d", &v);            if(v == 1) {                flag = true;            }        }        printf(flag ? "-1\n" : "1\n");    }    return 0;}

B
题意：给定一个大矩形和两个小矩形，问大矩形能否完全包含这两个小矩形，要求这两个小矩形不可以重复覆盖。

#include <iostream>#include <cstdlib>#include <cstdio>#include <algorithm>#include <string>#include <map>#include <cstring>#include <stack>#include <queue>#include <cmath>#include <vector>#include <set>using namespace std;typedef long long LL;typedef pair<int, int> pii;int main(){    int a1, b1, a2, b2, a3, b3;    while(scanf("%d%d", &a1, &b1) != EOF) {        scanf("%d%d", &a2, &b2);        scanf("%d%d", &a3, &b3);        bool flag = false;        if(a1 < b1) {            swap(a1, b1);        }        if(a2 < b2) {            swap(a2, b2);        }        if(a3 < b3) {            swap(a3, b3);        }        if(a2 + b3 <= a1 && max(b2, a3) <= b1) {            flag = true;        }        if(a2 + a3 <= a1 && max(b2, b3) <= b1) {            flag = true;        }        if(b2 + a3 <= a1 && max(a2, b3) <= b1) {            flag = true;        }        if(b2 + b3 <= a1 && max(a2, a3) <= b1) {            flag = true;        }        if(a2 + b3 <= b1 && max(b2, a3) <= a1) {            flag = true;        }        if(a2 + a3 <= b1 && max(b2, b3) <= a1) {            flag = true;        }        if(b2 + a3 <= b1 && max(a2, b3) <= a1) {            flag = true;        }        if(b2 + b3 <= b1 && max(a2, a3) <= a1) {            flag = true;        }        printf(flag ? "YES\n" : "NO\n");    }    return 0;}

C
题意：给定六边形的顺时针边长，已知每个角度均为120度。问你边长为1的正三角形有多少个。

我们每隔一个角添上一个新角，然后补成一个大的正三角形，最后去掉添上的三个角即可。

#include <iostream>#include <cstdlib>#include <cstdio>#include <algorithm>#include <string>#include <map>#include <cstring>#include <stack>#include <queue>#include <cmath>#include <vector>#include <set>using namespace std;typedef long long LL;typedef pair<int, int> pii;int a[7];int S(int n) {    return n * n;}int main(){    while(scanf("%d", &a[1]) != EOF) {        for(int i = 2; i <= 6; i++) {            scanf("%d", &a[i]);        }        printf("%d\n", S(a[1] + a[2] + a[3]) - S(a[1]) - S(a[3]) - S(a[5]));    }    return 0;}

D
题意：两个字符串匹配判定条件有两个，满足一个即可。
1、两个字符串相等；
2、严格从中间均分两份A1 A2 和 B1 B2，满足A1匹配B1且A2匹配B2 或者 A1匹配B2且A2匹配B1。

直接递归两个串显然不行，发现只要两个串按照上述方法组合后的最小字典序新串相同，那么它们一定是匹配。

#include <iostream>#include <cstdlib>#include <cstdio>#include <algorithm>#include <string>#include <map>#include <cstring>#include <stack>#include <queue>#include <cmath>#include <vector>#include <set>using namespace std;typedef long long LL;typedef pair<int, int> pii;const int MAXN = 1e5 + 10;const int MOD = 1e9 + 7;string Work(string a) {    int la = a.size(); if(la & 1) return a;    string a1 = a.substr(0, la >> 1), a2 = a.substr(la >> 1, la >> 1);    string b1 = Work(a1), b2 = Work(a2);    return b1 > b2 ? b2 + b1 : b1 + b2;}int main(){    string a, b;    while(cin >> a >> b) {        a = Work(a); b = Work(b);        if(a == b) {            cout << "YES" << endl;        }        else {            cout << "NO" << endl;        }    }    return 0;}

E

题意：每次只可以向下或者向右走，问你从(1, 1)到(n, n)的方案数，其中有m个格子是坏的。

dp[i]表示到达第i个坏格子且不经过前面所有坏格子的方案数。

#include <iostream>#include <cstdlib>#include <cstdio>#include <algorithm>#include <string>#include <map>#include <cstring>#include <stack>#include <queue>#include <cmath>#include <vector>#include <set>using namespace std;typedef long long LL;typedef pair<int, int> pii;const int MAXN = 1e5 + 10;const int MOD = 1e9 + 7;void add(LL &x, LL y) { x += y; if(x < 0) x += MOD; x %= MOD; }pii a[2010];LL fac[2 * MAXN];LL pow_mod(LL a, int n) {    LL ans = 1;    while(n) {        if(n & 1) {            ans = ans * a % MOD;        }        a = a * a % MOD;        n >>= 1;    }    return ans;}LL C(int n, int m) {    return fac[n] * pow_mod(fac[m], MOD - 2) % MOD * pow_mod(fac[n-m], MOD - 2) % MOD;}LL Solve(int x1, int y1, int x2, int y2) {    int x = x2 - x1 + 1; int y = y2 - y1 + 1;    return C(x + y - 2, x - 1);}LL dp[2010];int main(){    fac[0] = 1LL;    for(int i = 1; i <= 200000; i++) {        fac[i] = fac[i-1] * i % MOD;    }    int n, m, k;    while(scanf("%d%d%d", &n, &m, &k) != EOF) {        for(int i = 0; i < k; i++) {            scanf("%d%d", &a[i].first, &a[i].second);        }        sort(a, a + k); a[k] = pii(n, m);        for(int i = 0; i <= k; i++) {            dp[i] = 0; LL sum = 0;            for(int j = 0; j < i; j++) {                if(a[j].first <= a[i].first && a[j].second <= a[i].second) {                    add(sum, dp[j] * Solve(a[j].first, a[j].second, a[i].first, a[i].second));                }            }            add(dp[i], Solve(1, 1, a[i].first, a[i].second) - sum);        }        printf("%lld\n", dp[k]);    }    return 0;}