铺地毯
题解：比大小
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 ", Solve()); 19 }
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
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 ", Pri); 34 }
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
1 #include <cstdio> 2 #include <algorithm> 3 #define LL "%lld" 4 using namespace std; 5 6 typedef long long ll; 7 const int MAXN = 200000+10; 8 const ll INF = 1e12; 9 10 ll n, m, s, L, R, Mid, Pri, w[MAXN], l[MAXN], r[MAXN], v[MAXN], sum[MAXN], sumv[MAXN]; 11 12 inline bool check(int x){ 13 for (int i=1; i<=n; i++) 14 if (w[i]>=x) 15 sum[i] = sum[i-1]+1, 16 sumv[i] = sumv[i-1]+v[i]; 17 else 18 sum[i] = sum[i-1], 19 sumv[i] = sumv[i-1]; 20 21 ll cj = 0; 22 for (int i=1; i<=m; i++) 23 cj += (sum[r[i]]-sum[l[i]-1])*(sumv[r[i]]-sumv[l[i]-1]); 24 25 Pri = min(Pri, abs(cj-s)); 26 return s<=cj; 27 } 28 29 int main(){ 30 scanf(LL LL LL, &n, &m, &s), L = Pri = INF, R = -INF; 31 for (int i=1; i<=n; i++) 32 scanf(LL LL, w+i, v+i), 33 L = min(L, w[i]), R = max(R, w[i]); 34 for (int i=1; i<=m; i++) 35 scanf(LL LL, l+i, r+i); 36 37 while (L<=R){ 38 Mid = (L+R)>>1; 39 if (check(Mid)) L = Mid+1; 40 else R = Mid-1; 41 } 42 printf(LL, Pri); 43 }