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  • CF370 D Memory and Scores

    dp题 并运用了前缀和
    我看题目提示中有fft 我想了下感觉复杂度不过关还是未解

    #include<bits/stdc++.h>
    using namespace std;
    typedef long long ll;
    const int MAXN = 4e5+5;
    const int MOD = 1e9+7;
    const int ze = 2e5;
    ll dp[2][MAXN];
    ll sum[MAXN];
    
    int main(){
        int a,b,k,t;
        while(~scanf("%d %d %d %d",&a,&b,&k,&t)) {
            memset(dp,0,sizeof(dp));
            int fl= 1;
            dp[1][ze] = 1; sum[0] = 0;
            for(int i = 1; i <= t; ++i) {
                for(int j = 1; j < MAXN; ++j) {
                    sum[j] = (dp[fl][j]+sum[j-1]) %MOD;             
                }
    
                memset(dp[fl^1],0,sizeof(dp[fl^1]));
                for(int j = 1; j < MAXN; ++j) {
                    int t1 = max(0,j-k-1); int t2 = min(MAXN-1, j+k);
                    dp[fl^1][j] = (dp[fl^1][j]+sum[t2]-sum[t1]+MOD) %MOD;
                }
                fl ^= 1;
            }
            for(int i = 1; i < MAXN; ++i) {
                sum[i] = (dp[fl][i] + sum[i-1]) %MOD;
            }
            ll ans = 0;
            for(int i = 1; i < MAXN; ++i) {
                int tt = a-b+i;
                if(tt >= MAXN) tt = MAXN;
                if(tt < 1) tt = 1;
                ans = (ans + dp[fl][i]*sum[tt-1]%MOD)%MOD;
            }
            printf("%lld
    ",ans);
        }
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/Basasuya/p/8433754.html
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