标签：ash ace names print 下界 结束 预处理 air size
签到题，两种情况分别计算然后取个min
#include<vector>
#include<map>
#include<cstdio>
#define min(x, y) (x < y ? x : y)
#define max(x, y) (x < y ? y : x)
#define abs(x) (x < 0 ? - x : x)
#define Pair pair<int, int>
#define MP(x, y) make_pair(x, y)
using namespace std;
const int MAXN = 1e6 + 10, INF = 1e9 + 10;
inline int read() {
char c = getchar(); int x = 0, f = 1;
while(c < ‘0‘ || c > ‘9‘){if(c == ‘-‘) f = -1; c = getchar();}
while(c >= ‘0‘ && c <= ‘9‘) x = x * 10 + c - ‘0‘, c = getchar();
return x * f;
}
int N;
int a[MAXN];
int x = INF, y = INF, z = INF;
int main() {
#ifdef WIN32
freopen("a.in", "r", stdin);
#endif
N = read();
for(int i = 1; i <= N; i++) a[i] = read();
for(int i = 1; i <= N; i++) {
int opt = read();
if(opt == 1) x = min(x, a[i]);
else if(opt == 2) y = min(y, a[i]);
else if(opt == 3) z = min(z, a[i]);
}
printf("%d", min(x + y, z));
return 0;
}
不会QWQ....
首先一个很显然的性质就是当$w >8$或$w < -9$的时候是无解的
否则，我们令$D_1=x$，$D_N = x +Ｗ$，这样其它的数就可以任意取了，有$10^{N - 2}$种方案
然后把$x$的取值乘上
具体见代码吧
#include<cstdio>
#include<cstring>
#include<vector>
#include<map>
#include<cmath>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/hash_policy.hpp>
#define LL long long
using namespace __gnu_pbds;
using namespace std;
const LL MAXN = 1e6 + 10, INF = 1e9 + 10, mod = 1e9 + 7;
inline LL read() {
char c = getchar(); LL x = 0, f = 1;
while(c < ‘0‘ || c > ‘9‘){if(c == ‘-‘) f = -1; c = getchar();}
while(c >= ‘0‘ && c <= ‘9‘) x = x * 10 + c - ‘0‘, c = getchar();
return x * f;
}
LL T, N, W;
LL fastpow(LL a, LL p) {
LL base = 1;
while(p) {
if(p & 1) base = (base * a) % mod;
p >>= 1; a = (a * a ) % mod;
}
return base % mod;
}
main() {
#ifdef WIN32
freopen("a.in", "r", stdin);
//freopen("b.out", "w", stdout);
#endif
T = read();
while(T--) {
N = read(), W = read();
if(W > 8 || W < -9 ) {printf("0\n"); continue;}
printf("%lld\n", (W >= 0 ? (9 - W) * fastpow(10, N - 2) : (10 + W) * fastpow(10, N - 2) ) % mod);
}
}
考虑到值域很小，首先预处理出每个值的出现次数
然后枚举所有值，再枚举一个断点，根据这个数拆开后的两个数的出现次数算贡献
两个数相同的时候需要特判
存值域可以让每个数加上下界，也可以直接用cc_hash_table，可以卡过
#include<cstdio>
#include<cstring>
#include<vector>
#include<map>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/hash_policy.hpp>
#define min(x, y) (x < y ? x : y)
#define max(x, y) (x < y ? y : x)
#define abs(x) (x < 0 ? - x : x)
#define Pair pair<int, int>
#define MP(x, y) make_pair(x, y)
#define int long long
using namespace __gnu_pbds;
using namespace std;
const int MAXN = 1e6 + 10, INF = 1e9 + 10;
inline int read() {
char c = getchar(); int x = 0, f = 1;
while(c < ‘0‘ || c > ‘9‘){if(c == ‘-‘) f = -1; c = getchar();}
while(c >= ‘0‘ && c <= ‘9‘) x = x * 10 + c - ‘0‘, c = getchar();
return x * f;
}
int T, N;
int a[MAXN];
cc_hash_table<int,int>vis;
cc_hash_table<int,bool>happen;
main() {
#ifdef WIN32
freopen("a.in", "r", stdin);
#endif
T = read();
while(T--) {
N = read();
int mx = -INF, mi = INF; vis.clear(); happen.clear();
for(int i = 1; i <= N; i++) a[i] = read(), mx = max(a[i], mx), mi = min(a[i], mi);
for(int i = 1; i <= N; i++)
vis[a[i]]++;
int ans = 0;
for(int i = 1; i <= N; i++) {
int limit = 2 * a[i];
if(happen[limit]) continue;
for(int j = mi; j <= mx; j++) {
if(vis[j] && vis[limit - j]) {
if(j == (limit - j)) ans += (vis[j] - 1) * vis[j];
else ans += vis[j] * vis[limit - j];
}
}
happen[limit] = 1;
}
printf("%lld\n", ans / 2);
}
}
没啥可说的
大力特判，细节巨多
#include<cstdio>
#include<cstring>
#include<vector>
#include<map>
#include<cmath>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/hash_policy.hpp>
#define int long long
using namespace __gnu_pbds;
using namespace std;
const int MAXN = 1e6 + 10, INF = 1e9 + 10;
inline int read() {
char c = getchar(); int x = 0, f = 1;
while(c < ‘0‘ || c > ‘9‘){if(c == ‘-‘) f = -1; c = getchar();}
while(c >= ‘0‘ && c <= ‘9‘) x = x * 10 + c - ‘0‘, c = getchar();
return x * f;
}
int T, N;
char s[MAXN];
int a[MAXN], b[MAXN];
main() {
#ifdef WIN32
freopen("a.in", "r", stdin);
freopen("b.out", "w", stdout);
#endif
scanf("%d", &T);
while(T--) {
scanf("%s", s + 1); scanf("%d", &N);
int L = strlen(s + 1);
for(int i = 1; i <= L; i++)
a[i] = a[i - 1], b[i] = b[i - 1], s[i] == ‘a‘ ? a[i]++ : b[i]++;
int ans = 0;
if(a[L] > b[L]) {
for(int i = 1; i <= L; i++) {
if(a[i] > b[i]) ans = ans + N;
else if(a[i] == b[i]) ans = ans + N - 1;
else {
ans = ans + max((int)0, N - 1 - (b[i] - a[i]) / (a[L] - b[L]));
}
}
}
else if(a[L] == b[L]) {
for(int i = 1; i <= L; i++) {
if(a[i] > b[i]) ans = ans + N;
else continue;
}
}
else {
for(int i = 1; i <= L; i++) {
if(a[i] > b[i])
ans = ans + min(N, (a[i] - b[i]) / (b[L] - a[L]) + (bool)((a[i] - b[i]) % (b[L] - a[L])));
else continue;
}
}
printf("%lld\n", ans);
}
}
剩下的还没做
等APIO结束再说吧(估计也做不动了)
codechef April Challenge 2018 Div2(1-4)
标签：ash ace names print 下界 结束 预处理 air size
原文地址：https://www.cnblogs.com/zwfymqz/p/9026630.html
