NOIP2011 题解

 1 #include <cstdio>
 2 
 3 const int MAXN = 10000+10;
 4 
 5 int n, x, y, a[MAXN], b[MAXN], g[MAXN], k[MAXN];
 6 
 7 inline int Solve(){
 8     for (int i=n; i>0; i--)
 9         if (a[i]<=x && x<=a[i]+g[i] && b[i]<=y && y<=b[i]+k[i]) return i;
10     return -1;
11 }
12 
13 int main(){
14     scanf("%d", &n);
15     for (int i=1; i<=n; i++)
16         scanf("%d %d %d %d", &a[i], &b[i], &g[i], &k[i]);
17     scanf("%d %d", &x, &y);
18     printf("%d\n", Solve());
19 }

 1 #include <cstdio>
 2 #include <cstring>
 3 
 4 const int MAXK = 50+10;
 5 
 6 int n, k, p, color, cost, cj, Pri, legal[MAXK], tot[MAXK], last[MAXK];
 7 char c;
 8 
 9 inline int NextInt(){
10     int ret = 0;
11     do
12         c = getchar();
13     while (!(48<=c && c<=57));
14     
15     do
16         ret *= 10, ret += c-48, c = getchar();
17     while (48<=c && c<=57);
18 
19     return ret;
20 }
21 
22 int main(){
23     memset(tot, 0, sizeof(tot));
24     memset(legal, 0, sizeof(legal));
25 
26      n = NextInt(), k = NextInt(), p = NextInt(), Pri = 0;
27      for (int i=1; i<=n; i++){
28         color = NextInt(), cost = NextInt();
29         if (cost<=p) cj = i;
30         if (cj>=last[color]) legal[color] = tot[color];
31         Pri += legal[color] , tot[color] ++, last[color] = i;
32      }
33      printf("%d\n", Pri);
34 }

 1 #include <cstdio>
 2 #include <cstring>
 3 
 4 const int MAXN = 1000; 
 5 const int MOD = 10007;
 6 
 7 int a, b, k, m, n;
 8 long long c[MAXN+10][MAXN+10];
 9 
10 inline long long QuickPow(int a, int k){
11     long long base = a, ret = 1;
12     while (k){
13         if (k&1) ret = (ret*base)%MOD;
14         base = (base*base)%MOD, k >>= 1;
15     }
16     return ret%MOD;
17 }
18 
19 int main(){
20     memset(c, 0, sizeof(c));
21     for (register int i=0; i<=MAXN+1; i++)
22         c[i][i] = c[i][0] = 1;
23     for (register int i=2; i<=MAXN+1; i++)
24         for (register int j=1; j<i; j++)
25             c[i][j] = (c[i-1][j]+c[i-1][j-1])%MOD;
26 
27     scanf("%d %d %d %d %d", &a, &b, &k, &n, &m);
28     long long Pri = c[k][n];
29     Pri = (Pri*QuickPow(a,n))%MOD, Pri = (Pri*QuickPow(b,m))%MOD;
30     printf("%lld\n", Pri);
31 }