无题1

题解：

第一题：

这道题达到最终状态就停了，p是甲最终赢的概率，我们最后状态即p=1, 或p = 0;最初p = 0.5; 从0.5到1或0的增量都是0.5，甲还要赢x场， 乙还要赢y场，当x,y发生变动时，p改变，假设增量为q, 我们就应该投入 q/0.5 * 2 ^(2n - 1), （占总共钱的比列）；

那么这个q就可以通过上面的式子O（1）得了； 无限递推；

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const ll mod = 1e9 + 7;
const int M = 1e5 + 10;
ll v[M << 1],fac[M << 1], vfac[M << 1];

ll ksm(ll a, ll b){
    ll ret = 1;
    for(; b; b >>=1, a=a*a%mod)
        if(b & 1) ret=ret*a%mod;
    return ret;
}
ll ni(ll a){
    return ksm (a, mod - 2);
}
ll C(int a, int b){
    return fac[a] * vfac[b] % mod * vfac[a - b] % mod;
}

int main(){
    freopen("beijing.in","r",stdin);
    freopen("beijing.out","w",stdout);
    int n;
    scanf("%d", &n);
    v[1] = 1; fac[0] = vfac[0] = 1;
    for(ll i = 2; i <= 2 * n; i++){//推 i 的逆元
        v[i] = v[mod % i] * (mod - mod/i) % mod;
    }
    for(ll i = 1; i <= 2 * n; i++){
        fac[i] = fac[i - 1] * i % mod;
        vfac[i] = ni(fac[i]);
    }
    ll x = n, y = n;
    ll z = C(x + y - 2, x - 1) * 2 % mod;
    while(x > 0 && y > 0){
        printf("%lld
", z);
        z = z * v[x + y - 2] * 2 % mod;
        int opt; scanf("%d", &opt);
        if(!opt) z = z * (--x) % mod;
        else z = z * (--y) % mod;
    }
} 

第二题：打表找规律，发现就是加一个杨辉三角的*增量

空间开小了

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define RG register 
const int M = 2e4 + 10;
ll a[M], t[M], c[M], fac[M], vfac[M];
const ll mod = 1e9 + 7;
inline ll moc(ll a){
    return a >= mod ? a - mod : a;
}
int n, m, k; ll q;

ll ksm(ll a, ll b){
    ll ret = 1;
    for(; b; b >>=1, a=a*a%mod)
        if(b & 1) ret=ret*a%mod;
    return ret;
}
ll ni(ll a){
    return ksm (a, mod - 2);
}

void init(){
    fac[0] = vfac[0] = 1;
    for(ll i = 1; i <= 2*(n + 1); i++) fac[i] = fac[i - 1] * i % mod, vfac[i] = ni(fac[i]);
    for(int i = 1; i <= n + 1; i++) c[i] = fac[k - 2 + i] * vfac[k - 1] % mod * vfac[i - 1] % mod;
    
}



int main(){
    freopen("hongkong.in","r",stdin);
    freopen("hongkong.out","w",stdout);
    scanf("%d%d%d",&n, &m, &k);
    init();
    //for(int i = 1; i <= n; i++)printf("%lld ", c[i]);    
    int opt, x, y;
    while(m--){
        scanf("%d", &opt);
        if(!opt){
            scanf("%d%d", &x, &y);
            a[x] = moc(a[x] + y);
            for(int i = x; i <= n; i++) t[i] = moc(t[i] + c[i - x + 1] * y % mod);
        }
        else {
            scanf("%d", &x);
            printf("%lld
", t[x]);
        }
    }
}