codeforces简单思维题合集
1030C
题意是问一个数列能否划分为几段使得每段的和相等(段数至少为2)
必然有一段是从1开始的
所以答案必然是n-1个前缀和中的一个
O(n)枚举，O(n)检验
#include
using namespace std;
#define LL long long
#define db double
const int MAXN = 200;
const int LIM = 101;
const int INF = 0x3f3f3f3f;
template inline void read(T &x) {
int ch = getchar();
bool fg = false;
for (x = 0; !isdigit(ch); ch = getchar()) {
if (ch == '-') {
fg = true;
}
}
for (; isdigit(ch); ch = getchar()) {
x = x * 10 + ch - '0';
}
if (fg) {
x = -x;
}
}
int n, a[MAXN], sum[MAXN];
char s[MAXN];
bool check(int x, int k) {
int cur = k, tot = 1;
for(int i = k + 1; i <= n; i++) {
if(sum[i] - sum[cur] == x) {
cur = i, tot++;
while(sum[i] == sum[i + 1]) i++, cur++;
}
else while(sum[i + 1] == sum[i]) i++;
}
if(sum[n] - sum[cur] != x && cur != n) return 0;
if(tot > 1) return 1;
return 0;
}
signed main() {
read(n);
scanf("%s", s + 1);
if(n == 1) {
puts("NO");
return 0;
}
for(int i = 1; i <= n; i++) a[i] = (int) (s[i] - '0');
for(int i = 1; i <= n; i++) sum[i] = sum[i - 1] + a[i];
for(int i = 1; i <= n; i++) {
if(check(sum[i], i)) {
puts("YES");
return 0;
}
}
puts("NO");
return 0;
} 1031B
第一眼以为不可做
然鹅a,b的取值只有0,1,2,3，完全可以枚举t1然后递推检验
#include
#define LL long long
#define db double
using namespace std;
const int MAXN = 200010;
const int INF = 0x3f3f3f3f;
template inline void read(T &x) {
char c = getchar();
bool f = false;
for (x = 0; !isdigit(c); c = getchar()) {
if (c == '-') {
f = true;
}
}
for (; isdigit(c); c = getchar()) {
x = x * 10 + c - '0';
}
if (f) {
x = -x;
}
}
template inline bool CheckMax(T &a, const T &b) {
return a < b ? a = b, true : false;
}
template inline bool CheckMin(T &a, const T &b) {
return a > b ? a = b, true : false;
}
int n;
int a[MAXN], b[MAXN], t[MAXN];
int cnt;
#define ls lower_bound
#define us upper_bound
typedef pair pii;
typedef pair ppp;
map > f;
bool check(int T, int x) {
if(x == n) {
t[n] = T;
return 1;
}
t[x] = T;
for(int j = 0; j <= 3; j++) {
if((t[x] | j) == a[x] && (t[x] & j) == b[x]) {
return check(j, x + 1);
}
}
return 0;
}
signed main() {
read(n);
for(int i = 1; i < n; i++) read(a[i]);
for(int i = 1; i < n; i++) read(b[i]);
for(int i = 1; i < n; i++) {
if(a[i] < b[i]) {
puts("NO");
return 0;
}
}
for(int i = 0; i <= 3; i++) {
if(check(i, 1)) {
puts("YES");
for(int i = 1; i <= n; i++) printf("%d%c", t[i], i == n ? '\n' : ' ');
return 0;
}
}
puts("NO");
return 0;
} 1036D
给你两个数列，问能不能分别划分使得对应段的和相等
搞两个指针，戳戳戳就ojbk了
#include
using namespace std;
#define LL long long
#define db double
const int MAXN = 300300;
const int MAXE = 400400;
const int INF = 0x3f3f3f3f;
template inline void read(T &x) {
int ch = getchar();
bool fg = false;
for (x = 0; !isdigit(ch); ch = getchar()) {
if (ch == '-') {
fg = true;
}
}
for (; isdigit(ch); ch = getchar()) {
x = x * 10 + ch - '0';
}
if (fg) {
x = -x;
}
}
LL a[MAXN], b[MAXN];
int n, m;
signed main() {
read(n);
for(int i = 1; i <= n; i++) read(a[i]), a[i] += a[i - 1];
read(m);
for(int i = 1; i <= m; i++) read(b[i]), b[i] += b[i - 1];
if(a[n] != b[m]) puts("-1");
else {
int i = 1, j = 1, ans = 0;
while(i <= n && j <= m) {
while(a[i] < b[j] && i <= n) i++;
while(a[i] > b[j] && j <= m) j++;
if(a[i] == b[j] && i <= n && j <= m) {
i++, ans++;
}
}
printf("%d\n", ans);
}
return 0;
} 1036B
蜜汁找规律。。。
#include
using namespace std;
#define LL long long
#define db double
const int MAXN = 200200;
const int LIM = 1000000;
const int INF = 0x3f3f3f3f;
template inline void read(T &x) {
int ch = getchar();
bool fg = false;
for (x = 0; !isdigit(ch); ch = getchar()) {
if (ch == '-') {
fg = true;
}
}
for (; isdigit(ch); ch = getchar()) {
x = x * 10 + ch - '0';
}
if (fg) {
x = -x;
}
}
signed main() {
LL n, m, k, Q;
read(Q);
while(Q --) {
read(n), read(m), read(k);
if(max(n, m) > k) puts("-1");
else {
if((n + m) & 1) printf("%I64d\n", k - 1);
else if((k + m) & 1) printf("%I64d\n", k - 2);
else printf("%I64d\n", k);
}
}
return 0;
} 1060C
两个sigma并没有联系，所以可以单独考虑a和b，不必算出c
然后维护一下最大值就好了
#include
#define LL long long
#define db double
using namespace std;
const int MAXN = 2002;
const int MAXE = 400400;
const int INF = 0x3f3f3f3f;
template inline void CheckMax(T &A, T B) {
A < B ? A = B : A;
}
template inline void CheckMin(T &A, T B) {
A > B ? A = B : A;
}
template inline void read(T &x) {
int c = getchar();
bool f = false;
for (x = 0; !isdigit(c); c = getchar()) {
if (c == '-') {
f = true;
}
}
for (; isdigit(c); c = getchar()) {
x = x * 10 + c - '0';
}
if (f) {
x = -x;
}
}
int n, m;
LL a[MAXN], b[MAXN], X;
LL asum[MAXN], bsum[MAXN];
signed main() {
read(n), read(m);
for(int i = 1; i <= n; i++) read(a[i]);
for(int i = 1; i <= m; i++) read(b[i]);
read(X);
for(int i = 1; i <= n; i++) a[i] += a[i - 1];
for(int j = 1; j <= m; j++) b[j] += b[j - 1];
memset(asum, 0x7f, sizeof(asum));
memset(bsum, 0x7f, sizeof(bsum));
for(int len = 1; len <= n || len <= m; len++) {
if(len <= n) {
for(int i = 1; i <= n - len + 1; i++) {
CheckMin(asum[len], a[i + len - 1] - a[i - 1]);
}
}
if(len <= m) {
for(int i = 1; i <= m - len + 1; i++) {
CheckMin(bsum[len], b[i + len - 1] - b[i - 1]);
}
}
}
int ans = 0;
for(int i = 1; i <= n; i++) {
for(int j = 1; j <= m; j++) {
if(asum[i] * bsum[j] <= X)
CheckMax(ans, i * j);
}
}
printf("%d\n", ans);
return 0;
}