NOIP2011 题解

铺地毯

　　题解：比大小

 #include <cstdio>

 const int MAXN = +;

 int n, x, y, a[MAXN], b[MAXN], g[MAXN], k[MAXN];

 inline int Solve(){
     for (int i=n; i>; i--)
         if (a[i]<=x && x<=a[i]+g[i] && b[i]<=y && y<=b[i]+k[i]) return i;
     return -;
 }

 int main(){
     scanf("%d", &n);
     for (int i=; i<=n; i++)
         scanf("%d %d %d %d", &a[i], &b[i], &g[i], &k[i]);
     scanf("%d %d", &x, &y);
     printf("%d\n", Solve());
 }

carpet.cpp

选择客栈:

　　题解：模拟。cj表示最近的一个可行的咖啡馆，tot[i]表示之前颜色i出现过的次数legal[i]表示颜色i在cj之前出现过的次数，last[i]表示颜色i最后出现的编号

 #include <cstdio>
 #include <cstring>

 const int MAXK = +;

 int n, k, p, color, cost, cj, Pri, legal[MAXK], tot[MAXK], last[MAXK];
 char c;

 inline int NextInt(){
     int ret = ;
     do
         c = getchar();
     while (!(<=c && c<=));

     do
         ret *= , ret += c-, c = getchar();
     while (<=c && c<=);

     return ret;
 }

 int main(){
     memset(tot, , sizeof(tot));
     memset(legal, , sizeof(legal));

      n = NextInt(), k = NextInt(), p = NextInt(), Pri = ;
      for (int i=; i<=n; i++){
         color = NextInt(), cost = NextInt();
         if (cost<=p) cj = i;
         if (cj>=last[color]) legal[color] = tot[color];
         Pri += legal[color] , tot[color] ++, last[color] = i;
      }
      printf("%d\n", Pri);
 }

hotel.cpp

计算系数：

　　题解：组合数+快速幂。之前一直wa，直到把一些变量变成long long，然后多mod几次，就过了...

factor.cpp

聪明的质检员：

　　题解：二分

 #include <cstdio>
 #include <algorithm>
 #define LL "%lld"
 using namespace std;

 typedef long long ll;
 const int MAXN = +;
 const ll INF = 1e12;

 ll n, m, s, L, R, Mid, Pri, w[MAXN], l[MAXN], r[MAXN], v[MAXN], sum[MAXN], sumv[MAXN];

 inline bool check(int x){
     for (int i=; i<=n; i++)
         if (w[i]>=x)
             sum[i] = sum[i-]+,
             sumv[i] = sumv[i-]+v[i];
         else
             sum[i] = sum[i-],
             sumv[i] = sumv[i-];

     ll cj = ;
     for (int i=; i<=m; i++)
         cj += (sum[r[i]]-sum[l[i]-])*(sumv[r[i]]-sumv[l[i]-]);

     Pri = min(Pri, abs(cj-s));
     return s<=cj;
 }

 int main(){
     scanf(LL LL LL, &n, &m, &s), L = Pri = INF, R = -INF;
     for (int i=; i<=n; i++)
         scanf(LL LL, w+i, v+i),
         L = min(L, w[i]), R = max(R, w[i]);
     for (int i=; i<=m; i++)
         scanf(LL LL, l+i, r+i);

     while (L<=R){
         Mid = (L+R)>>;
         if (check(Mid)) L = Mid+;
         else R = Mid-;
     }
     printf(LL, Pri);
 }

check.cpp