NOIP2000普及组

第一题 计算器的改良

大模拟，好好处理字符串。

 

第二题 税收与补贴

略过。

 

第三题 乘积最大

区间动归，记得高精度。

 

第四题 单词接龙

带字符处理的深搜。

 

代码

第一题

 

#include <stdio.h>

#include <ctype.h>

#include <string.h>

#define MAXN 300

char s[MAXN],c;

double xx,yy,x;

bool jia,havenum;

int wh;

int main() {

  scanf("%s",s);

  jia = 1;

  for (int i = 0;i < strlen(s);++i)

    if (s[i] == '=') {

      wh = i;

      break;

    }

  for (int i = 0;i < wh;++i)

    if (isdigit(s[i])) {

      havenum = 1;

      x = x * 10 + s[i] - '0';

    } else {

        if (isalpha(s[i])) {

          c = s[i];

          if ((!havenum) && (!x))

            x = 1;

          if (jia)

            xx += x;

          else

            xx -= x;

          havenum = 0;

          x = 0;

        } else {

            if (havenum)

              if (jia)

                yy -= x;

              else

               yy += x;

            if (s[i] == '+')

              jia = 1;

            else

              jia = 0;

            x = 0;

          }

      }

  if (havenum)

    if (jia)

      yy -= x;

    else

     yy += x;

  x = 0;

  havenum = 0;

  jia = 1;

  for (int i = wh + 1;i < strlen(s);++i)

    if (isdigit(s[i])) {

      havenum = 1;

      x = x * 10 + s[i] - '0';

    } else {

        if (isalpha(s[i])) {

          c = s[i];

          if ((!havenum) && (!x))

            x = 1;

          if (jia)

            xx -= x;

          else

            xx += x;

          havenum = 0;

          x = 0;

        } else {

            if (havenum)

              if (jia)

                yy += x;

              else

                yy -= x;

            if (s[i] == '+')

              jia = 1;

            else

              jia = 0;

            x = 0;

          }

      }

  if (havenum)

    if (jia)

      yy += x;

    else

      yy -= x;

  printf("%c=%.3lf\n",c,yy / xx);

  return 0;

}

第三题

#include <stdio.h>

#include <string.h>

#define MAXL 310

#define MAXN 50

#define MAXM 10

struct ss {

  int g[MAXL];

} f[MAXN][MAXM],a[MAXN][MAXN];

char s[MAXN];

int n,m;

ss bigger(ss a,ss b) {

  if (a.g[0] < b.g[0])

    return b;

  if (a.g[0] > b.g[0])

    return a;

  for (int i = a.g[0];i > 0;--i) {

    if (a.g[i] < b.g[i])

      return b;

    if (a.g[i] > b.g[i])

      return a;

  }

  return a;

}

ss cheng(ss a,ss b) {

  ss c;

  memset(c.g,0,sizeof(c.g));

  for (int i = 1;i <= a.g[0];++i) {

    int x = 0;

    for (int j = 1;j <= b.g[0];++j) {

      c.g[i + j - 1] +=  a.g[i] * b.g[j] + x;

      x = c.g[i + j - 1] / 10;

      c.g[i + j - 1] %= 10;

    }

    if (x)

      c.g[i + b.g[0]] += x;

  }

  c.g[0] = a.g[0] + b.g[0];

  while ((c.g[0] > 1) && (!c.g[c.g[0]]))

    --c.g[0];

  return c;

}

void out(ss a) {

  for (int i = a.g[0];i > 0;--i)

    printf("%d",a.g[i]);

}

int main() {

  scanf("%d%d",&n,&m);

  ++m;

  scanf("%s",s);

  for (int i = n;i > 0;--i)

    s[i] = s[i - 1];

  for (int i = n;i > 0;--i) {

    a[i][i].g[0] = 1;

    a[i][i].g[1] = s[i] - '0';

    for (int j = i - 1;j > 0;--j) {

      a[j][i] = a[j + 1][i];

      a[j][i].g[++a[j][i].g[0]] = s[j] - '0';

    }

  }

  for (int i = 1;i <= n;++i)

    f[i][1] = a[1][i];

  for (int i = 1;i <= n;++i)

    for (int j = 2;j <= m && j <= i;++j)

      for (int k = j - 1;k < i;++k)

        f[i][j] = bigger(f[i][j],cheng(f[k][j - 1],a[k + 1][i]));

  out(f[n][m]);

  return 0;

}

第四题

#include <stdio.h>

#include <string.h>

#define MAXN 21

struct ss {

  char s[300];

  int len,time;

} a[MAXN];

int n,ans;

char te[30000010];

void find(int x) {

  if (x > ans)

    ans = x;

  for (int i = 1;i <= n;++i)

    if (a[i].time < 2)

      for (int j = 1;j < a[i].len;++j) {

        if (x - j < 0)

          break;

        bool can = 1;

        for (int k = 1;k <= j;++k)

          if (te[x - j + k] != a[i].s[k]) {

            can = 0;

            break;

          }

        if (can) {

          ++a[i].time;

          for (int k = j + 1;k <= a[i].len;++k)

            te[x + k - j] = a[i].s[k];

          find(x + a[i].len - j);

          --a[i].time;

        }

      }

}

int main() {

  scanf("%d",&n);

  for (int i = 1;i <= n;++i) {

    scanf("%s",a[i].s);

    a[i].len = strlen(a[i].s);

    for (int j = a[i].len;j > 0;--j)

      a[i].s[j] = a[i].s[j - 1];

  }

  for (int i = 1;i <= n;++i) {

    for (int j = 1;j <= a[i].len;++j)

      te[j] = a[i].s[j];

    ++a[i].time;

    find(a[i].len);

  }

  printf("%d\n",ans);

  return 0;

}