[2018CCPC吉林赛区（重现赛）- 感谢北华大学] 补题记录 躁起来

1007 High Priestess

埃及分数

1008 Lovers

线段树维护取膜意义下的区间s和。

每个区间保存前缀lazy和后缀lazy。

#include 
using namespace std;
#define pb push_back
#define fi first
#define se second
#define debug(x) cerr<<#x << " := " << x << endl;
#define bug cerr<<"-----------------------"<#define FOR(a, b, c) for(int a = b; a <= c; ++ a)

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair pll;


template<class T> void _R(T &x) { cin >> x; }
void _R(int &x) { scanf("%d", &x); }
void _R(ll &x) { scanf("%lld", &x); }
void _R(double &x) { scanf("%lf", &x); }
void _R(char &x) { scanf(" %c", &x); }
void _R(char *x) { scanf("%s", x); }
void R() {}
template<class T, class... U> void R(T &head, U &... tail) { _R(head); R(tail...); }


template
inline T read(T&x){
    x=0;int f=0;char ch=getchar();
    while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar();
    while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x=f?-x:x;
}

const int inf = 0x3f3f3f3f;

const int mod = 1e9+7;

/**********showtime************/
            const int maxn = 1e5+9;
            ll sum[maxn<<2];
            ll lazylen[maxn<<2],lazyback[maxn<<2],lazypre[maxn<<2];
            ll sumten[maxn<<2];
            char str[10];

            void build(int le, int ri, int rt) {
                sum[rt] = 0;
                lazylen[rt] = lazyback[rt] = lazypre[rt] = 0;
                sumten[rt] = 1;
                if(le == ri) {
                    return;
                }
                int mid = (le + ri) >> 1;
                build(le, mid, rt<<1);
                build(mid+1, ri, rt<<1|1);
                sumten[rt] = sumten[rt<<1] + sumten[rt<<1|1];
            }

            ll ten[maxn];
            void pushdown(int le, int ri, int rt){
                    int mid = (le + ri) >> 1;
                    sumten[rt<<1] = ten[lazylen[rt]] * sumten[rt<<1]%mod;
                    sum[rt<<1] = ((sum[rt<<1]*ten[lazylen[rt]]%mod + sumten[rt<<1]*lazypre[rt]%mod )%mod+ 1ll*(mid-le+1)*lazyback[rt] % mod)%mod;
                    sumten[rt<<1] = ten[lazylen[rt]] * sumten[rt<<1]%mod;

                    sumten[rt<<1|1] = ten[lazylen[rt]]*sumten[rt<<1|1]%mod;
                    sum[rt<<1|1] = ((sum[rt<<1|1]*ten[lazylen[rt]]%mod + sumten[rt<<1|1]*lazypre[rt]%mod )%mod+ 1ll*(ri-mid)*lazyback[rt] % mod)%mod;
                    sumten[rt<<1|1] = ten[lazylen[rt]]*sumten[rt<<1|1]%mod;

                    lazypre[rt<<1] = (lazypre[rt] * ten[lazylen[rt<<1]]%mod + lazypre[rt<<1] )% mod;
                    lazyback[rt<<1] = ((lazyback[rt<<1] * ten[lazylen[rt]])%mod + lazyback[rt]) % mod;
                    lazylen[rt<<1] =  lazylen[rt<<1] + lazylen[rt];

                    lazypre[rt<<1|1] = (lazypre[rt] * ten[lazylen[rt<<1|1]]%mod + lazypre[rt<<1|1])% mod;
                    lazyback[rt<<1|1] = (lazyback[rt<<1|1] * ten[lazylen[rt]]%mod + lazyback[rt]) % mod;
                    lazylen[rt<<1|1] =  lazylen[rt<<1|1] + lazylen[rt];

                    lazylen[rt] = 0;
                    lazyback[rt] = 0;
                    lazypre[rt] = 0;
            }

            void pushup(int rt) {
                sum[rt] = (sum[rt<<1] + sum[rt<<1|1])%mod;
                sumten[rt] = (sumten[rt<<1] + sumten[rt<<1|1]) % mod;
            }

            void update(int L, int R, int b, int le, int ri, int rt) {
                if(le >= L && ri <= R) {
                    sumten[rt] = 1ll*10*sumten[rt]%mod;
                    sum[rt] = ((1ll*sum[rt]*10%mod + 1ll*sumten[rt]*b%mod )%mod + 1ll*(ri-le+1)*b % mod)%mod;
                    sumten[rt] = 1ll*10*sumten[rt]%mod;

                    lazypre[rt] = (1ll* b * ten[lazylen[rt]]%mod + lazypre[rt]) % mod;
                    lazyback[rt] = (1ll*lazyback[rt] * 10 %mod + b)%mod;
                    lazylen[rt]++;
                    return;
                }
                if(lazylen[rt]) pushdown(le, ri, rt);
                int mid = (le + ri) >> 1;
                if(mid >= L) update(L, R, b, le, mid, rt<<1);
                if(mid < R) update(L, R, b, mid+1, ri, rt<<1|1);
                pushup(rt);
            }

            ll query(int L, int R, int le, int ri, int rt) {
                if(le >= L && ri <= R) {
                    return sum[rt];
                }
                if(lazylen[rt]) pushdown(le, ri, rt);
                int mid = (le + ri) >> 1;
                ll res = 0;
                if(mid >= L) res = (res + query(L, R, le, mid, rt<<1) )% mod;
                if(mid < R) res = (res + query(L, R, mid+1, ri, rt<<1|1)) % mod;
                pushup(rt);
                return res;
            }
int main(){
//            freopen("data.in", "r", stdin);
            ten[0] = 1;
            for(int i=1; i1] * 10 % mod;
            int T;  scanf("%d", &T);
            int cas = 0;
            while(T--) {
                int n,m;
                scanf("%d%d", &n, &m);
                build(1, n, 1);
                printf("Case %d:\n", ++cas);
                while(m--) {
                    scanf("%s", str);
                    if(str[0] == 'w') {
                        int le, ri, b;
                        scanf("%d%d%d", &le, &ri, &b);
                        update(le, ri, b, 1, n, 1);
                    }
                    else {
                        int le, ri;
                        scanf("%d%d", &le, &ri);
                        printf("%lld\n", query(le, ri, 1, n , 1));
                    }
                }
            }
            return 0;
}

View Code

 

1010 Wheel of Fortune

转化为判定问题，是否存在一种对于球的完备匹配，且

匹配次数最多的颜色数量不超过 ⌊n⌋。 2

1011 The Magician

硬核模拟

1012 The Hanged Man

树链剖分

普通树形背包复杂度 O(nm2)，不可取。 考虑在 dfs 序上做 01 背包。

 

转载于:https://www.cnblogs.com/ckxkexing/p/11196487.html