Android應用基礎學習記錄

標籤：比賽總結

福州大學第十一屆程式設計競賽
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Tag	Pro.ID	Problem Title	Ratio(AC/Submit)
	1	大王叫我來巡山呐	80.31%(261/325)
	2	防守陣地 I	23.48%(193/822)
	3	shadow	13.23%(97/733)
	4	花生的序列	12.07%(21/174)
	5	防守陣地 II	14.98%(68/454)
	6	巡了南山我巡北山	0.00%(0/45)
	7	Nostop	20.79%(21/101)
	8	卷福的難題	10.00%(1/10)

        

               本來打算當作組隊賽來做的，結果隊友有足球賽，還有一個有黨校。。只剩下我一個人單挑了。 @[email protected]   還好不是很難，不然就被虐翻了。。整體來說題目比較簡單，應該本來就是作為個人賽準備的。。

               A題籤到題

#include<algorithm>#include<iostream>#include<cstring>#include<vector>#include<cstdio>#include<cmath>#include<queue>#include<stack>#include<map>#include<set>#define LL long long#define CLR(a, b) memset(a, b, sizeof(a))using namespace std;const int maxn = 5;const int INF = 0x3f3f3f3f;int main(){    int n;    while(scanf("%d", &n) != EOF)    {        int ans = n / 7 * 2;        if(n % 7 == 6) ans ++;        printf("%d\n", ans);    }}

                  B題觀察式子轉移，就可以很容易發現端倪了。。

#include<algorithm>#include<iostream>#include<cstring>#include<vector>#include<cstdio>#include<cmath>#include<queue>#include<stack>#include<map>#include<set>#define LL long long#define CLR(a, b) memset(a, b, sizeof(a))using namespace std;const int maxn = 1001000;const int INF = 0x3f3f3f3f;int a[maxn];int main(){    int n, m;    while(scanf("%d%d", &n, &m) != EOF)    {        int ans = 0, tot = 0, sum = 0;        for(int i = 1; i <= m; i ++)        {            scanf("%d", &a[i]);            sum += a[i];            tot += i*a[i];        }        ans = tot;        for(int i = m + 1; i <= n; i ++)        {            scanf("%d", &a[i]);            tot -= sum;            sum -= a[i - m];            sum += a[i];            tot += a[i] * m;            ans = max(ans, tot);        }        printf("%d\n", ans);    }}

             C題是比較簡單的樹dp吧，只需要看孩子和本身有沒有士兵，有的話當前點肯定會被消滅。OJ比較奇特，遞迴會爆棧，用vector會逾時。。不幸的是兩種情況我都遇到了。。

#include<algorithm>#include<iostream>#include<cstring>#include<vector>#include<cstdio>#include<cmath>#include<queue>#include<stack>#include<map>#include<set>#define LL long long#define MP make_pair#define CLR(a, b) memset(a, b, sizeof(a))using namespace std;const int maxn = 100100;const int INF = 0x3f3f3f3f;int arm[maxn];bool hav[maxn];struct Edge{    int u, v;}E[maxn*2];int fir[maxn], nxt[maxn*2], tot = 0;LL d[maxn];void AddEdge(int u, int v){    E[tot].u = u, E[tot].v = v;    nxt[tot] = fir[u], fir[u] = tot ++;}LL dfs(){    stack<pair<int, int> > S;    CLR(d, 0);    S.push(MP(1, 0));    int cnt = 0;    while(!S.empty())    {        int u = S.top().first, fa = S.top().second;        if(fir[u] == -1)        {            if(hav[u]) hav[fa] = 1, d[fa] += d[u] + arm[u];            S.pop();        }        else        {            int v = E[fir[u]].v;            fir[u] = nxt[fir[u]];            if(v == fa) continue;            S.push(MP(v, u));        }    }    return d[1];}int main(){    int n, k;    while(scanf("%d%d", &n, &k) != EOF)    {        for(int i = 1; i <= n; i ++)        {            scanf("%d", &arm[i]);        }        CLR(hav, 0);CLR(fir, -1); tot = 0;        for(int i = 0; i < k; i ++)        {            int u; scanf("%d", &u);            hav[u] = 1;        }        for(int i = 1; i < n; i ++)        {            int u, v;            scanf("%d%d", &u, &v);            AddEdge(u, v);            AddEdge(v, u);        }        printf("%lld\n", dfs());    }}

            D題算是一道dp吧，dp[i][j]，表示到第i為為止，有j個屬於第一個串，因為串的特殊性（WBWBWB），所以很容易判斷當前位置的字母滿足的情況。比如當前j為奇數，那麼j後面肯定只能接B。。。就這樣。。。

#include<algorithm>#include<iostream>#include<cstring>#include<vector>#include<cstdio>#include<cmath>#include<queue>#include<stack>#include<map>#include<set>#define LL long long#define CLR(a, b) memset(a, b, sizeof(a))using namespace std;const int maxn = 10100;const int INF = 0x3f3f3f3f;const int MOD = 1000000007;char ch[maxn];int dp[2][3300];int main(){    int T, n;    scanf("%d", &T);    while(T --)    {        scanf("%d%s", &n, ch);        CLR(dp, 0);        dp[0][0] = 1;        for(int i = 0, j = 1; i < n * 2; i ++, j ++)        {            CLR(dp[j&1], 0);            if(ch[i] == ‘B‘)            {                for(int k = 0; k <= n; k ++)                {                    if(k&1) dp[j&1][k+1] = (dp[j&1][k+1] + dp[i&1][k]) % MOD;                    if((i-k)&1) dp[j&1][k] = (dp[j&1][k] + dp[i&1][k]) % MOD;                }            }            else            {                for(int k = 0; k <= n; k ++)                {                    if((k&1) == 0) dp[j&1][k+1] = (dp[j&1][k+1] + dp[i&1][k]) % MOD;                    if(((i-k)&1) == 0) dp[j&1][k] = (dp[j&1][k] + dp[i&1][k]) % MOD;                }            }        }        printf("%d\n", dp[0][n]);    }}

               E裸線段樹區間更新，區間查詢，只要你不想那麼多，就可以直接A掉。

#include<algorithm>#include<iostream>#include<cstring>#include<vector>#include<cstdio>#include<cmath>#include<queue>#include<stack>#include<map>#include<set>#define LL long long#define CLR(a, b) memset(a, b, sizeof(a))#define lson l , m , rt << 1#define rson m + 1 , r , rt << 1 | 1using namespace std;const int maxn = 100100;const int INF = 0x3f3f3f3f;int add[maxn<<2];int sum[maxn<<2];void PushUp(int rt){    sum[rt] = sum[rt<<1] + sum[rt<<1|1];}void PushDown(int rt,int m){    if (add[rt])    {        add[rt<<1] += add[rt];        add[rt<<1|1] += add[rt];        sum[rt<<1] += add[rt] * (m - (m >> 1));        sum[rt<<1|1] += add[rt] * (m >> 1);        add[rt] = 0;    }}void build(int l,int r,int rt){    add[rt] = 0;    if (l == r)    {        scanf("%d",&sum[rt]);        return ;    }    int m = (l + r) >> 1;    build(lson);    build(rson);    PushUp(rt);}void update(int L,int R,int c,int l,int r,int rt){    if (L <= l && r <= R)    {        add[rt] += c;        sum[rt] += c * (r - l + 1);        return ;    }    PushDown(rt , r - l + 1);    int m = (l + r) >> 1;    if (L <= m) update(L , R , c , lson);    if (m < R) update(L , R , c , rson);    PushUp(rt);}int query(int L,int R,int l,int r,int rt){    if (L <= l && r <= R)    {        return sum[rt];    }    PushDown(rt , r - l + 1);    int m = (l + r) >> 1;    LL ret = 0;    if (L <= m) ret += query(L , R , lson);    if (m < R) ret += query(L , R , rson);    return ret;}int main(){    int n, m, q;    while(scanf("%d%d%d", &n, &m, &q) != EOF)    {        build(1, n, 1);        for(int i = 0; i < q; i ++)        {            int x; scanf("%d", &x);            printf("%d\n", query(x, x + m - 1, 1, n, 1));            update(x, x + m - 1, -1, 1, n, 1);        }    }}

                F應該是搜尋吧。。。感覺無從下手，罪過。。。大致思路應該是判斷是否是子串，然後在進行加字元，copy操作吧。。。我猜的。。

                G題這種題第一次見到，感覺自己YY出來還是很高興的！就是用矩陣來表示走K步之後的各個點對到達所需要的最小花費。這樣的話，我們就可以用初始矩陣（表示的是1步到達的情況）不斷更新，來獲得走2步，走3步的到達情況。。這樣的話思路就很明確了。因為題目資料比較大，我們直接用快速冪更新就可以了。（應該很容易想到這些更新是滿足結合率的吧）。我還是覺得看代碼清楚些。。這個OJ很個性，用lld，wa了無數次，改成cin，cout才過。。。=.=!!

#include<algorithm>#include<iostream>#include<cstring>#include<vector>#include<cstdio>#include<cmath>#include<queue>#include<stack>#include<map>#include<set>#define LL long long#define MP make_pair#define CLR(a, b) memset(a, b, sizeof(a))using namespace std;const int maxn = 55;const LL INF = 1ll << 60;struct Matrix{    LL m[maxn][maxn];}sol, pre;int n;Matrix Add(Matrix a, Matrix b){    Matrix ret;    for(int i = 1; i <= n; i ++)    {        for(int j = 1; j <= n; j ++)        {            ret.m[i][j] = INF;            LL& tmp = ret.m[i][j];            for(int k = 1; k <= n; k ++)            {                tmp = min(tmp, a.m[i][k] + b.m[k][j]);            }        }    }    return ret;}Matrix Mul(Matrix a, int n){    Matrix ret;    bool flag = false;    while(n)    {        if(n & 1)        {            if(flag) ret = Add(ret, a);            else ret = a;            flag = true;        }        a = Add(a, a);        n >>= 1;    }    return ret;}void pt(Matrix sol){    for(int i = 1; i <= n; i ++)    {        for(int j = 1; j <= n; j ++)            printf("%lld ", sol.m[i][j]);            puts("");    }}int main(){    int h, k, T;    scanf("%d", &T);    while(T --)    {        scanf("%d%d%d", &n, &h, &k);        for(int i = 1; i <= n; i ++)        {            for(int j = 1; j <= n; j ++)                sol.m[i][j] = INF;        }        for(int i = 0; i < h; i ++)        {            int u, v;LL c;            scanf("%d%d", &u, &v); cin >> c;            sol.m[u][v] = min(sol.m[u][v], c);        }        sol = Mul(sol, k);        if(sol.m[1][n] == INF) puts("-1");        else cout << sol.m[1][n] << endl;        //printf("%lld\n", sol.m[1][n]);    }}

                 H題完全沒想法，最後絕殺的那個人真厲害，ORZ～～

Android應用基礎學習記錄