2017"百度之星"程序设计大赛

1001.Arithmetic of Bomb

直接模拟求出最后的串就好了。

#include <iostream>#include <cstdio>#include <cstring>#include <queue>#include <algorithm>#include <string>using namespace std;char s[1005];long long ans = 0;const int mod = 1000000007;struct Node{    char s[1005];    int cnt;};Node gao2(int &pos){    pos--;    Node temp;    temp.cnt = s[pos] - '0';    pos--;    while (s[pos] < '0' || s[pos] > '9') pos--;    int t = 0;    for (; s[pos] != '('; pos--){        temp.s[t++] = s[pos];    }    temp.s[t] = 0;    return temp;}void gao(long long x){    x %= mod;    ans = ans + x;    ans %= mod;}int main(){    int t;    scanf("%d", &t);    while (t--)    {        scanf("%s", s);        int len = strlen(s);        ans = 0;        long long pre = 1;        for (int i = len - 1; i >= 0; i--){            if (s[i] == ')'){                Node now = gao2(i);                while (now.cnt--){                    int len1 = strlen(now.s);                    for (int j = 0; j < len1; j++){                        gao((now.s[j] - '0') * pre);                        pre *= 10;                        pre %= mod;                    }                }            }            else{                gao((s[i] - '0') * pre);                pre *= 10;                pre %= mod;            }        }        printf("%I64d\n", ans);    }}

1003.Pokémon GO

解法1：dp
解法2：找规律。。。

#include <iostream>#include <cstdio>#include <cstring>#include <queue>#include <algorithm>#include <string>using namespace std;const long long mod = 1e9 + 7;//long long biao[] = { 1, 6, 12, 26, 48, 86, 148, 250, 416, 686, 1124, 1834, 2984, 4846, 7860, 12738, 20632, 33406, 54076, 87522, 141640, 229206, 370892, 600146, 971088, 1571286, 2542428, 4113770, 6656256, 10770086, 17426404, 28196554, 45623024, 73819646, 119442740, 193262458 };long long qpow(long long x, long long a) {    long long ret = 1;    while (a) {        if (a & 1) ret = (ret*x) % mod;        x = (x*x) % mod;        a >>= 1;    }    return ret;}long long num[11000];int main(){    num[0] = 1;    num[1] = 1;    num[2] = 6;    num[3] = 12;    for (int i = 4; i <= 10000; i++) {        num[i] =          (( ( ( ( (8 * i - 31) * num[i - 1] % mod + mod ) % mod - (4 * i - 19) * num[i - 2] % mod + mod) % mod ) - (3 * i - 10) * num[i - 3] % mod + mod) % mod + (2 * i - 10) * num[i - 4] + mod) % mod * qpow(3 * i - 11, mod - 2) % mod;    }    int T;    cin >> T;    while (T--) {        long long N;        cin >> N;        long long ans = num[N] * qpow(2, N) % mod;        cout << ans << endl;    }}

1005.Valley Numer

数位dp。
dp[p][pre][isdown][lim][st]表示进行到第p位，前一位是pre，是否是下降的，是否有限制，是否之前都是前导0时的种类数。

#include <iostream>#include <cstdio>#include <cstring>#include <queue>#include <algorithm>#include <string>using namespace std;const long long MOD = 1e9 + 7;long long dp[110][11][2][2][2];int len, num[110];char str[110];long long dfs(int p, int pre, int isdown, int lim, int st){    if (p == len + 1)return !st;    if (dp[p][pre][isdown][lim][st] != -1 && !lim)return dp[p][pre][isdown][lim][st];    long long temp = 0;    int r = lim ? num[p] : 9;    for (int i = 0; i <= r; i++)    {        if (isdown)         {            if (i > pre) {                temp += dfs(p + 1, i, 0, lim&&i == r, 0) % MOD;            }            else {                if (st && i == 0) {                    temp += dfs(p + 1, 10, 1, lim&&i == r, st) % MOD;                }                else {                    temp += dfs(p + 1, i, 1, lim&&i == r, 0) % MOD;                }            }        }        else if (i >= pre) {            temp += dfs(p + 1, i, 0, lim&&i == r, 0) % MOD;        }        temp = temp % MOD;    }    return dp[p][pre][isdown][lim][st] = temp % MOD;}int main(){    int T;    scanf("%d\n", &T);    while (T--)    {        scanf("%s", str);        len = strlen(str);        for (int i = 0; i<len; i++)            num[i + 1] = str[i] - '0';        memset(dp, -1, sizeof(dp));        printf("%I64d\n", dfs(1, 10, 1, 1, 1));    }    return 0;}