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;
}