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  • 比赛-NOIP2015 普及组复赛 (Nov, 2016)

    先贴上代码,迟一些会附上文字讲解。

     1 #include <cstdio>
     2 int k, ans;
     3 
     4 int main()
     5 {
     6     freopen("coin.in", "r", stdin);
     7     freopen("coin.out", "w", stdout);
     8     
     9     int day = 0, i = 0;
    10     scanf("%d", &k);
    11     do {
    12         if ((day += ++i) > k) {
    13             ans -= (day - k - i) * i;
    14             break;
    15         }
    16         ans += i * i;
    17     } while (true);
    18     
    19     printf("%d
    ", ans);
    20     
    21     fclose(stdin);
    22     fclose(stdout);
    23     
    24     return 0;
    25 }
    coin
     1 #include <cstdio>
     2 
     3 const int MAX_N = 105, MAX_M = 105;
     4 
     5 int n, m;
     6 int nxt[8][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}, 
     7                 {-1, 1}, {-1, -1}, {1, 1}, {1, -1}};
     8 bool G[MAX_N][MAX_M];
     9 
    10 void input()
    11 {
    12     int i, j;
    13     char tt;
    14     scanf("%d%d", &n, &m);
    15     for (i = 1; i <= n; ++i)
    16         for (j = 1; j <= m; ++j) {
    17             while ((tt = getchar()) == '
    ' || tt == ' ');
    18             G[i][j] = tt == '*';
    19         }
    20     return ;
    21 }
    22 
    23 void solve()
    24 {
    25     int i, j, k, ans;
    26     for (i = 1; i <= n; ++i) {
    27         for (j = 1; j <= m; ++j) {
    28             if (G[i][j]) {
    29                 putchar('*');
    30                 continue;
    31             }
    32             ans = 0;
    33             for (k = 0; k < 8; ++k)
    34                 ans += G[i + nxt[k][0]][j + nxt[k][1]];
    35             printf("%d", ans);
    36         }
    37         putchar('
    ');
    38     }
    39     return ;
    40 }
    41 
    42 int main()
    43 {
    44     freopen("mine.in", "r", stdin);
    45     freopen("mine.out", "w", stdout);
    46     
    47     input();
    48     solve();
    49     
    50     fclose(stdin);
    51     fclose(stdout);
    52     return 0;
    53 }
    mine
     1 #include <cstdio>
     2 
     3 const int MAX_NM = 100005, MOD = 10007;
     4 
     5 int n, m, ans;
     6 int N[MAX_NM], C[MAX_NM];
     7 int k[2][MAX_NM], sum[2][MAX_NM];
     8 
     9 inline void getnum(int &num)
    10 {
    11     char tt;
    12     bool flag = false;
    13     while (((tt = getchar()) < '0' || tt > '9') && tt != '-');
    14     if (tt == '-') {
    15         flag = true;
    16         num = 0;
    17     }
    18     else {
    19         num = tt - '0';
    20     }
    21     while ((tt = getchar()) >= '0' && tt <= '9')
    22         num = num * 10 + tt - '0';
    23     if (flag)
    24         num = -num;
    25     return ;
    26 }
    27 
    28 inline void add_mod(int &a, int b)
    29 {
    30     a = (a + b % MOD) % MOD;
    31     return ;
    32 }
    33 
    34 int main()
    35 {
    36     freopen("sum.in", "r", stdin);
    37     freopen("sum.out", "w", stdout);
    38     
    39     int i, tmp;
    40     getnum(n);
    41     getnum(m);
    42     for (i = 1; i <= n; ++i) {
    43         getnum(N[i]);
    44         N[i] %= MOD;
    45     }
    46     for (i = 1; i <= n; ++i) {
    47         getnum(C[i]);
    48         add_mod(k[i & 1][C[i]], 1);
    49         add_mod(sum[i & 1][C[i]], N[i]);
    50     }
    51     for (i = 1; i <= n; ++i)
    52         add_mod(ans, i % MOD * (N[i] * (k[i & 1][C[i]] - 2) % MOD + sum[i & 1][C[i]]) % MOD);
    53     
    54     printf("%d
    ", ans);
    55     
    56     fclose(stdin);
    57     fclose(stdout);
    58     
    59     return 0;
    60 }
    sum
      1 #include <cstdio>
      2 #include <algorithm>
      3 using namespace std;
      4 
      5 const int MAX_N = 100005;
      6 
      7 int n, ans, top_l;
      8 int S[MAX_N], A[MAX_N];
      9 int l[MAX_N];//max heap
     10 char s[25];
     11 
     12 struct house {
     13     int num;
     14     int w;
     15 } W[MAX_N];
     16 
     17 
     18 inline void getnum(int &num)
     19 {
     20     char tt;
     21     bool flag = false;
     22     while (((tt = getchar()) < '0' || tt > '9') && tt != '-');
     23     if (tt == '-') {
     24         flag = true;
     25         num = 0;
     26     }
     27     else {
     28         num = tt - '0';
     29     }
     30     while ((tt = getchar()) >= '0' && tt <= '9')
     31         num = num * 10 + tt - '0';
     32     if (flag)
     33         num = -num;
     34     return ;
     35 }
     36 
     37 void putnum(int num)
     38 {
     39     int top = 0;
     40     if (num < 0) {
     41         putchar('-');
     42         num = -num;
     43     }
     44     do
     45         s[++top] = num % 10 + '0';
     46     while (num /= 10);
     47     while (top)
     48         putchar(s[top--]);
     49     putchar('
    ');
     50     return ;
     51 }
     52 
     53 bool my_cmp(struct house a, struct house b)
     54 {
     55     return a.w > b.w;
     56 }
     57 
     58 void shift_up(int k, int *heap, int top)
     59 {
     60     int dad = k >> 1, tmp = heap[k];
     61     while (dad >= 1) {
     62         if (tmp > heap[dad]) {
     63             heap[k] = heap[dad];
     64             k = dad;
     65             dad >>= 1;
     66         } else {
     67             break;
     68         }
     69     }
     70     heap[k] = tmp;
     71     return ;
     72 }
     73 
     74 void shift_dn(int k, int *heap, int top)
     75 {
     76     int son = k << 1, tmp = heap[k];
     77     while (son <= top) {
     78         if (son < top && heap[son] < heap[son + 1])
     79             ++son;
     80         if (tmp < heap[son]) {
     81             heap[k] = heap[son];
     82             k = son;
     83             son <<= 1;
     84         } else {
     85             break;
     86         }
     87     }
     88     heap[k] = tmp;
     89     return ;
     90 }
     91 
     92 int main()
     93 {
     94     freopen("salesman.in", "r", stdin);
     95     freopen("salesman.out", "w", stdout);
     96     
     97     
     98     int i, x, dis = 0, p = 1;
     99     bool flag = false;
    100     
    101     getnum(n);
    102     for (i = 1; i <= n; ++i)
    103         getnum(S[i]);
    104     for (i = 1; i <= n; ++i) {
    105         getnum(A[i]);
    106         W[i].w = A[i] + (S[i] << 1);
    107         W[i].num = i;
    108     }
    109     
    110     sort(W + 1, W + n + 1, my_cmp);
    111     
    112     for (x = 1; x <= n; ++x) {
    113         int tmp;
    114         if (p > n || (top_l && l[1] > W[p].w - (S[dis] << 1))) {
    115             ans += l[1];
    116             l[1] = l[top_l];
    117             shift_dn(1, l, --top_l);
    118         } else {
    119             ans += W[p].w - (S[dis] << 1);
    120             dis = W[p].num;
    121             while (++p <= n && S[W[p].num] < S[dis]) {
    122                 l[++top_l] = A[W[p].num];
    123                 shift_up(top_l, l, top_l);
    124             }
    125         }
    126         putnum(ans);
    127     }
    128     
    129     fclose(stdin);
    130     fclose(stdout);
    131     return 0;
    132 }
    salesman
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  • 原文地址:https://www.cnblogs.com/ghcred/p/6061590.html
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