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715a3ccb41
...
5b884a0621
@ -1,108 +0,0 @@
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#include <cstdio>
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#include <cctype>
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const int N = 10;
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int n, a[N], s[N], ans;
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template <typename T>
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void read(T& x) {
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x = 0;
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char c = getchar();
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int f = 1;
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while (!isdigit(c)) {
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if (c == '-') f = -1;
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c = getchar();
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}
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while (isdigit(c)) {
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x = x * 10 + (c - '0');
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c = getchar();
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}
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x *= f;
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}
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void dfs(int x, int y) {
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// 当前得分已经大于最终得分
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if (s[x] > a[x]) return;
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// 如果以后的比赛全赢也小于最后的分数
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if (s[x] + (n - y + 1) * 3 < a[x]) return;
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if (x == n && s[x] == a[x]) {
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ans++;
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return;
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}
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if (y == n) {
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int t = a[x] - s[x];
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// 最后一场比赛无法凑出 2 分
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switch (t) {
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case 0: {
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s[y] += 3;
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dfs(x + 1, x + 2);
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s[y] -= 3;
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break;
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}
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case 1: {
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s[x] += 1;
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s[y] += 1;
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dfs(x + 1, x + 2);
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s[x] -= 1;
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s[y] -= 1;
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break;
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}
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case 3: {
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s[x] += 3;
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dfs(x + 1, x + 2);
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s[x] -= 3;
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break;
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}
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}
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return;
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}
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// x 胜
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s[x] += 3;
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s[y] += 0;
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dfs(x, y + 1);
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s[x] -= 3;
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s[y] -= 0;
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// 平局
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s[x] += 1;
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s[y] += 1;
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dfs(x, y + 1);
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s[x] -= 1;
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s[y] -= 1;
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// y 胜
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s[x] += 0;
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s[y] += 3;
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dfs(x, y + 1);
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s[x] -= 0;
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s[y] -= 3;
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}
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int main() {
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read(n);
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for (int i = 1; i <= n; i++) {
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read(a[i]);
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}
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dfs(1, 2);
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printf("%d\n", ans);
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return 0;
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}
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@ -1,108 +0,0 @@
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#include <cstdio>
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#include <cctype>
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const int N = 10;
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int n, a[N], s[N], ans;
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template <typename T>
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void read(T& x) {
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x = 0;
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char c = getchar();
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int f = 1;
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while (!isdigit(c)) {
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if (c == '-') f = -1;
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c = getchar();
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}
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while (isdigit(c)) {
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x = x * 10 + (c - '0');
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c = getchar();
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}
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x *= f;
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}
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void dfs(int x, int y) {
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// 当前得分已经大于最终得分
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if (s[x] > a[x]) return;
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// 如果以后的比赛全赢也小于最后的分数
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if (s[x] + (n - y + 1) * 3 < a[x]) return;
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if (x == n && s[x] == a[x]) {
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ans++;
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return;
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}
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if (y == n) {
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int t = a[x] - s[x];
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// 最后一场比赛无法凑出 2 分
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switch (t) {
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case 0: {
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s[y] += 3;
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dfs(x + 1, x + 2);
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s[y] -= 3;
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break;
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}
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case 1: {
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s[x] += 1;
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s[y] += 1;
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dfs(x + 1, x + 2);
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s[x] -= 1;
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s[y] -= 1;
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break;
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}
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case 3: {
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s[x] += 3;
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dfs(x + 1, x + 2);
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s[x] -= 3;
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break;
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}
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}
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return;
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}
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// x 胜
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s[x] += 3;
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s[y] += 0;
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dfs(x, y + 1);
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s[x] -= 3;
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s[y] -= 0;
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// 平局
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s[x] += 1;
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s[y] += 1;
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dfs(x, y + 1);
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s[x] -= 1;
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s[y] -= 1;
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// y 胜
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s[x] += 0;
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s[y] += 3;
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dfs(x, y + 1);
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s[x] -= 0;
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s[y] -= 3;
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}
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int main() {
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read(n);
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for (int i = 1; i <= n; i++) {
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read(a[i]);
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}
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dfs(1, 2);
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printf("%d\n", ans);
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return 0;
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}
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